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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 151. |
Write a relation between `Delta G and Q` and define the meaning of each term and answer the following: (a) Why a reaction proceeds forward when `Q lt K` and no net reaction occurs when Q = K. (b) Explain the effect of increase in pressure in terms of reaction Q, for the reaction : `CO(g) + 3 H_(2) (g) hArr CH_(4)(g) + H_(2) O (g)`. |
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Answer» `Delta G = DeltaG^(@)` + RT ln Q where `Delta G =` Change in free energy as the reaction proceeds `Delta G^(@) = ` Standard free energy change Q = Reaction quotient R = Gas constant and T= temperature in K (a) As `Delta G^(@) = - ` RT ln K `:. Delta G = - RT` ln K + RT ln Q = RT ln `(Q)/(K)` If `Q lt K, Delta G` will be -ve. Reaction will proceed in the forward direction. If `Q=K, DeltaG=0`. Reaction is in equilibrium and there is no net reaction . (b) `K_(c)=([CH_(4)][H_(2)O])/([CO][H_(2)]^(3))`. If pressure is increased, volume decreases. Suppose presure is doubled, volume will be halved. Hence, molar concentrations will be doubled. Now `Q_(c) = ({2[CH_(4)]}{2[H_(2)O]})/({2[CO]}{2[H_(2)]}^(3))=(1[CH_(4)][H_(2)O])/(4[CO][H_(2)]^(3))=(1)/(4) K_(c)` This, `Q_(c)` is less than `K_(c)`. Hence, to re-establish equilibrium , `Q_(c) ` will tend to increase , i.e., equilibrium will shift in the forward direction. |
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| 152. |
The reaction which proceeds in the forward direction is.A. `SnCl_4+Hg_2Cl_2rarrSnCl_2+2HgCl_2`B. `2Cul+I_+4K^(+)rarr 2Cu^(2+)+4Kl`C. `NH_3+H_2O+NaClrarr NH_4Cl+NaOH`D. `Fe_2O_3+6HClrarr2FeCl_3+3H_2O` |
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Answer» Correct Answer - D Hydrated ferric chloride `(FeCl_3-6H_2O)` is obtained by the action of hydrochloride acid on ferric oxide. (1) Stannous chloride is a powerful reducing agent. When an excess of stannous chloride solution is added to mercuric chloride solution, a white precipitate of mercurous chloride appears which slowly turns grey as further reduction to mercury occurs: `SnCl_2+2HgCl_2rarrHg_2Cl_2darr+SnCl_4` `Hg_2Cl_2+SnCl_2rarr2Hg_2darr+SnCl_4` (2) Cuprous iodie is precipitated as a buff solid when potassium iodide solution is added to the solution of cupric ions (e.g. copper sulphate solution): `2CuSO_4+2Klrarr2CuIdarr+I_2+2K_2SO_4` This reaction is used for estimating copper salts volumetrically by titrating the liberated iodine with standard sodium thiosulphate solution. (3) `NaOH` is a stronger base than `NH_3`, and thus, diplaces it from its salt: `NH_4Cl+NaOHrarrNaCl+underset(NH_3+H_2O)underset(darr)(NH_4OH)` |
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| 153. |
Finding equilibrium concentrations: A mixture of `0.50` mol `H_2` and `0.50` mol `I_2` is placed in a `1.00L` stainless steel container at `400^@C`. The equilibrium constant `K_c` for the reaction `H_2(g)+I_2(g)hArr 2HI(g)` is `54.3` at this temperature. Calculate the equilibrium concentrations of `H_2`, `I_2`, and `HI`. |
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Answer» Strategy: First we find the initial concentrations. Then we write the reaction summary and represent the equilibrium concentrations algebraically. Finally, we substitute the algebric representations of equilibrium concentrations into the `K_c` expression and find the equilibrium concentrations. Solution: Step: The initial concentrations are `CH_2=(n_(H_2))/(V_L)=(0.50mol)/(1.00L)=0.50 mol^-1` `C_(I_2)=(n_(I_2))/(V_L)=(0.50 mol)/(1.00L)=0.50 mol L^-1` The reaction can only proceed to the light because only reactants are present. The reaction summary includes the values, or symbols for the values, of (i) initial concentrations, (ii) changes in concentrations, and (iii) concentrations at equilibrium. Step 2: According to the stoichiometry of the reaction, 1 mol `H_2` reacts with 1 mol `I_2` to produce 2 mol `HI`. Let x be the decrease in concentration `(mol L^(-1))` of either `H_2` or `I_2` at equilibrium. Then the equilibrium concentrations of `HI` must be `2x`. Summarizing the changes in concentrations, we get `{:(,H_(2)(g)+ I_(2)(g)hArr 2HI(g)),("Initial (M)":,"0.50 0.50 0.00"),("Change (M)":,"-x -x +2x"),("Equilibrium",bar((0.50-x) (0.50) " " 2x)):}` Step 2: According to the law of chemical equilibrium, the equilibrium constant is given by `K_c=(C_(HI)^2)/(C_(H_2)C_(I_2))` Now, `K_c` is known but concentrations are not. But the equilibrium concentrations have all been expressed in terms of a single variable x. We substitute the equilibrium concentrations (not the initial ones) into the `K_c` expressions and solve for x. Substituting, we get `54.3=((2x)^2)/((0.50-x)(0.50-x))` `=((2x)^2)/((0.50-x)^2)` Taking the square root of both sides, we get `7.37=(2x)/((0.50-x))` `(7.37)(0.50-x)=2x` `3.685-7.37x=2x` `3.685=2x+7.37x=9.37x` `x=(3.685)/(9.37)=0.393M` Step: Now, we know the value of x. So the equilibrium concentrations are `C_(H_2)=(0.50-0.393)M=0.107M` `C_(I_2)=(0.50-0.393)M=0.107M` `C_(HI)=2xx0.393M=0.786M` Step 4: To check our answers, we use the equilibrium concentrations to calculate `Q_c` and verify that its value is equal to `K_c`. `Q_c=(C_(HI)^2)/(C_(H_2)C_(I_2))=((0.786)^2)/((0.107)(0.107))=54.0` |
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| 154. |
0.049 g of `H_(2)SO_(4)` is dissolved per litre of the given solution . Calculate the pH of the solution. |
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Answer» Correct Answer - `3.0` `[H_(2)SO_(4)]=(0.049)/(98) "mol" L^(-1)=5xx10^(-4)M`. As `H_(2)SO_(4)rarr2H^(+)+SO_(4)^(2-), [H^(+)]=2xx(5xx10^(-4))=10^(-3)M :. pH = 3` |
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| 155. |
How can you predict the following stages of a reaction by comparing the value of `K_(c) and Q_(c)`? (i) Net reaction proceeds in the forward direction. (ii) Net reaction proceeds in the backward direction. (iii) No net reaction occurs. |
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Answer» When `Q_(c) lt K_(c)`, net reaction proceeds in the forward direction. (ii) When `Q_(c) gt K_(c)`, net reaction proceeds in the backward direction. (iii) When `Q_(c) = K_(c)`, reaction is in equilibrium and no net reaction occurs. |
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| 156. |
The equilibrium constant `K_c` for the reaction `H_2(g)+I_2(g)hArr 2HI(g)` is `54.3` at `400^@C`. If the initial concentrations of `H_2`, `I_2` and `HI` are `0.00623M`, `0.00414M`, and `0.0224M`, respectively, calculate the concentrations of these species at equilibrium. |
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Answer» The only difference is that now we have non-stoichiometric amounts of reactants. Also, find the value of `Q_c` to decide the direction of reaction. Solution: Let x be the decrease in concentration `(mol L^(-1))` for `H_2` and `I_2` at equilibrium. Then, according to the stoichiometry of the reaction, the increase in concentration for `HI` must be `2x`. Summarize the changes n concentrations as follows: `{:(," "H_(2) (g)" " +" "I_(2)(g)" " hArr 2HI(g)),("Initial (M)", " 0.00623 0.00414 0.0224"),("Change (M)", " -x -x +2x"),("Equilibrium (M)", bar((0.00623-x) (0.00414-x) (0.0224+2x))):}` Note that we have considered the net reaction in the forward direction because `Q_c(19.45)` is less than `K_c(54.3)`. Step 2: Writing the equilibrium constant expression `K_c=(C_(HI)^2)/(C_(H_2)C_(I_2))` Substituting, we get `K_c=((0.0224+2x)^2)/((0.0063-x)(0.00414-x))` It is not possible to solve this equation by the square root shortcut as the left side of this equation is not a perfect square, because the starting concentrations of `H_2` and `I_2` are unequal. Instead, we can arrange this quardratic equation into the standard form by carrying out the multiplications: `54.3(0.00623-x)(0.00414-x)=(0.0224+2x^2)` `54.3(2.58xx10^5-0.0104x+x^2)=5.02xx10^-4+0.0896x+4x^2` Collecting terms, we get `50.3x^2-0.65x+8.98xx10^-4=0` This is a quardratic equation of the form `ax^2+bx+c=0`, which can be solved by the quardratic formula `x=(-b+-sqrt(b^2-4ac))/(2a)` In this case, we have `a=50.3, b=-0.654`, and `c=8.98xx10^-4` Substituting these values gives `x=(0.654+-sqrt((-0.654)^2-4(50.3)(8.98xx10^-4)))/(2xx50.3)` `implies x=0.0114M` or `x=0.00156M` The first solution is physically impossible since the amounts of `H_2` and `I_2` reacted would be more than those originally present. The second solution gives the correct answer. Note that in solving a quardratic equation of this type, one answer is always physically impossible. Thus, the choice of which value to use for x is easy to make. Step 3: The equilibrium concentrations are `C_(H_2)=(0.00623-0.00156)M=0.00467M` `C_(I_2)=(0.00414-0.00156)M=0.00258M` `C_(HI)=(0.0224+2xx0.00156)M=0.0255M` |
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| 157. |
Which of the followings are Lewis acids: `H_(2)O, BF_(3), H^(o+)` and `NH_(4)`? |
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Answer» Correct Answer - `BF_(3),H^(+),NH_(4^(+))` Lewis acids are those acids which can accept a pair of electrons. For example, `BF_(3)`, `H^(+)`, and `NH_(4)^(+)` are Lewis acids. |
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| 158. |
The ionization constant of nitrous acid is `4.5xx10^(-4)`. Calculate the `pH` of `0.04 M` sodium nitrite solution and also its degree of hydrolysis. |
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Answer» Correct Answer - `pH = 7.97`. Degree of hydrolysis `= 2.36 xx 10^(-5)` `"NaNO"_(2)` is the salt of a strong base (NaOH) and a weak acid `"HNO"_(2)`. `NO_(2)^(-)+H_(2)OleftrightarrowHNO_(2)+OH^(-)` `K_(h) = ([HNO_(2)][OH^(-)])/([NO_(2)^(-)])` `rArr (K_(w))/(K_(a)) = (10^(-14))/(4.5 xx 10^(-4)) = .22 xx 10^(10)` Now, If x moles of the salt undergo hydrolysis, then the concentration of various species present in the solution will be: `[NO_(2)^(-)] = .04-x , 0.04` `[HNO_(2)] = x` `[OH^(-)] = x` `K_(b) = (x^(2))/(0.04) = 0.22 xx 10^(-10)` `x^(2) = .0088 xx 10^(-10)` `x = .093 xx 10^(-5)` `:. [OH^(-)] = 0.093 xx 10^(-5) M` `[HO^(+)] = (10^(-14))/(.093 xx 10^(-5)) = 10.75 xx 10^(-9) M` `rArr pH = - "log" (10.75 xx 10^(-19))` `= 7.96` Therefore, degree of hydrolysis `= (x)/(0.04) = (.093 xx 10^(-5))/(.04) = 2.325 xx 10^(-5)` |
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| 159. |
0.1 M HCl and 0.1 M `H_(2)SO_(4)`, each of volume 2 ml are mixed and the volume is made up to 6 ml by adding 2 ml of 0.01 N NaCl solution. The pH of the resulting mixture isA. `1.17`B. `1.0`C. `0.3`D. log 2 - log 3 |
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Answer» Correct Answer - B 2 ml of 0.1 M HCl = 0.2 millimole of HCl = 0.2 millimole of `H^(+)` ion 2 ml of 0.1 M `H_(2)SO_(4)= 0 .2` millimole of `H_(2)SO_(4)` = 0.4 millimole of `H^(+)` ion Total `H^(+)` ions = 0.6 millimole Total volume = 6 ml `:. [H^(+)]=(0.6)/(6)M=0.1 M = 10^(-1)M :. pH = 1`. |
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| 160. |
How can you prdict the following stages of a reaction by comparing the value of `K_c`and Q_c` ? (i) Net reaction proceeds in the toward direction.(ii) Net recation proceeds in the backward dirction. (iii) No net rection occurs. |
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Answer» We know that `Q_c` is concentration quotient or reaction quotient. While `K_c` is the equilibrium constant. (i) Net reaction proceeds in the forward dirction if `Q_c lt K_c` (ii) Net reaction proceeds in the forward dirction if `Q_c gt K_c` (iii) No net reaction occurs if `Q_c =K_c`. |
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| 161. |
A `0.02 M` solution of pyridinium hydrochloride has `pH=3.44`. Calculate the ionization constant of pyridine. |
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Answer» Correct Answer - `K_(b) = 1.5 xx 10^(-9)` pH = 3.44 We know that, `pH=-log[H^(+)]` `therefore[H^(+)]=3.63xx10^(-4)` Then, `K_(h)=((3.63xx10^(-4))^(2))/(0.02)" "(because"concentration = 0.02M")` `rArrK_(h)=6.6xx10^(-6)` Now, `K_(h)=(K_(w))/(K_(a))` `rArrK_(a)=(K_(w))/(K_(h))=(10^(-14))/(6.6xx10^(-6))` `=1.51xx10^(-9)` |
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| 162. |
Assertion. In case of polyprotic acids, first ionization constant in lowest. Reason. The removal of first proton is most difficult. Further ionization becomes easier.A. If both assertion and reason are true, and reason is the true explanation of the assertion.B. If both assertion and reason are true, but reason is not the true explanation of the assertion.C. If aasertion is true, but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - D Correct A. In case of polyprotic acids, first ionization constant is highest. Correct R. First proton is lost easily. Further, loss of `H^(+)` ion becomes difficult because it remains attracted by the negative ion. |
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| 163. |
Assertion. pH of neutral solution is always 7. Reason. pH of a solution does not depend upon temperature.A. If both assertion and reason are true, and reason is the true explanation of the assertion.B. If both assertion and reason are true, but reason is not the true explanation of the assertion.C. If aasertion is true, but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - D Correct A. pH of a neutral solution is 7 only at 298 K. Correct R. pH of a solution changes with temperature. |
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| 164. |
The molar solubility of silver sulphate is `1.5xx10^(-2)mol L^(-1)`. The solubility product of the salt will beA. `2.25xx10^(-4)`B. `1.4xx10^(-5)`C. `1.7xx10^(-6)`D. `3.0xx10^(-3)` |
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Answer» Correct Answer - B Solubility equilibrium equation: `Ag_(2)SO_(4)(s)hArr2Ag^(+)(aq.)+SO_(4)^(2-)(aq.)` 1 mol of `Ag_(2)SO_(4)` produces 2 mol of `Ag^(+)` and 1 mol of `SO_(4)^(2-)` in solution. Therefore, when `1.5xx10^(-2)` mol `Ag_(2)SO_(4)` is dissolved in 1 L of solution, the concentrations are `C_(Ag^(+))= 2(1.5xx10^(-2)M)` `= 3.0xx10^(-2)M` `C_(SO_(4)^(2-))= 1.5xx10^(-2)M` Solubility product: `K_(sp)=C_(Ag^(+))^(2)C_(SO_(4)^(2-))` `= (3.0xx10^(-2))^(2)(1.5xx10^(-2))` `= 1.4xx10^(-5)` |
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| 165. |
Solubility if `M_(2)S` type salt is `3.5xx10^(-6)`, then find out its solubility productA. `1.7 xx 10^(-6)`B. `3.4 xx 10^(-16)`C. `1.7 xx 10^(-16)`D. `6.8 xx 10^(-12)` |
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Answer» Correct Answer - C `M_(2)S hArr underset(2S)(2M^(2+)) +underset(S)(S^(2-))` `4S^(3) =K_(sp) =3.5 xx 10^(-6)` `S=((3.5 xx 10^(-6))/(4))^(1//3) = 1.7 xx 10^(-16)` |
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| 166. |
Given : ` 2 N_(2) O (g) hArr 2 N_(2) (g) + O_(2) (g) , K= 3*5 xx 10^(33)` ` 2 NO _(2) (g) hArr N_(2) (g) + 2 O_(2) (g) , K= 6*7 xx 10^(16)` ` 2 NO (g) hArr N_(2) (g) + O_(2) (g) , K= 2*2 xx 10^(30)` ` 2 N_(2)O_(5) (g) hArr 2 N_(2) (g) + 5 O_(2) (g), K+1*2 xx 10^(34)` Which oxide of nitrogen is most stable ?A. `N_(2)O`B. `NO_(2)`C. NOD. `N_(2) O_(5)` |
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Answer» Correct Answer - B Lower the value of K, less is the dissociation and hence grater is the stability. |
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| 167. |
The value of `K_(c)` for the reaction, `2 HI (g) hArr H_(2)g + I_(2) (g) ` is `1xx10^(-4)` At a given time, the composition of reaction mixture is `[HI] = 2xx10^(-5) "mol" and [I_(2)]=1xx10^(-5) ` mol In which direction will the reaction proceed ? |
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Answer» Reaction quotient (Q) `= ([H_(2)][I_(2)])/([HI]^(2))=((10^(-5))(10^(-5)))/((2xx10^(-5))^(2))=(1)/(4) = 0.25 =2.5xx10^(-1)` As `Q gt K_(c)`, the reaction will proceed in the reverse direction. |
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| 168. |
The equilibrium constant value `K_(p)` for the equilibrium : ` H_(2) (g) + I_(2) hArr 2 HI (g) ` changes withA. total pressureB. temperatureC. catalystD. amount of `H_(2) and I_(2)` present |
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Answer» Correct Answer - B The value of equilibrium constant of a reaction depends only on temperature and soes not depend upon concentration , pressure or presence of catalyst . |
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| 169. |
the value of equilibrium constant for the reaction `HI(g) hArr 1//2 H_(2) (g) +1//2 I_(2) (g) " is " 8.0` The equilibrium constant for the reaction `H_(2) (g) +I_(2) (g) hArr 2HI(g)` will beA. 16B. `1//8`C. `1//16`D. `1//64` |
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Answer» Correct Answer - D `HI(g) hArr 1//2 H_(2) (g) + 1//2I_(2) (g)` `K= [[H_(2)]^(1//2) [I_(2)]^ (1//2)]/[[HI]] =8` `H_(2)(g) +I_(2) (g) hArr 2HI(g)` `K =[[H]^(2)]/[[H_(2)][I_(2)]] =((1)/(8))^(2) =(1)/(64)` |
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| 170. |
The dissociation equilibrium of a gas `AB_(2)` can be represented as : `2AB_(2) hArr 2AB(g) +B_(2)(g)` The degree of dissocaition x is very small as compared to 1. The expression which relates the degree of dissociation (x) with equilibrium constant `(K_(p))` and total pressure (p) is :A. `(2 K_(p) //P)^(1//2)`B. `(K_(p)//P)`C. `(2K_(p)//P)`D. `(2 K_(p)//P)^(1//3)` |
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Answer» Correct Answer - D `2AB_(2) (g) hArr 2AB (g) + B_(2)(g)` `"Initial mol:"" "2" "0" "0` `"Eqm mol:"" "2(1-x)" "2x" "x` No of moles at equilibrium. `=2(1 -x) + 2x +x =(2 + x) " mol "` `K_(p) = (p^(2) AB xx pB)/(p^(2) AB_(2))` `=(((2x)/(2+x))^(2) xx ((x)/(2+x)))/(((2(1-x))/((2+x)))^(2))=((4x^(3))/((2+x))xxP)/(4(1-x)^(2))` `K_(p) =((4x^(3) xx P)/(2))/(4) [2+xx~~2 " and " 1-x~~1]` `K_(p) =(x^(3)p)/(2) " or " x^(3) =(2Kp)/(P)` `x= ((2Kp)/(P))^(1//3)` |
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| 171. |
The vapour pressure of a liquid in a closed vesselA. depandes upon the amount of the liquid taken sB. Keeps on increasing continously as more and more liquid evaporatesC. has a constant value depending only on the nature of the liquidD. had a constant value at constant temperature |
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Answer» Correct Answer - D The vapour pressure of a liquid has a constant value at constant temperature . |
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| 172. |
Write the unit of equlibrium constant `(K_(c))` for the given reaction. `BaCO_(3)(s)hArrBaO(s)+CO_(2)(g)` |
| Answer» `K_(c)=[CO_(2)]=mol L^(-1)` | |
| 173. |
For `4NH_(3)(g)+5O_(2)(g)hArr4NO(g)+6H_(2)O(g),` write the expression of `K_(c)` |
| Answer» `K_(c)=([NO]^(4)[H_(2)O]^(6))/([NH_(3)]^(4)[O_(2)]^(5))` | |
| 174. |
Equal volumes of three acid solutions of `pH 3, 4` and `5` are mixed in a vessel. What will be the `H^(+)` ion concentration in the mixture? |
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Answer» For first solution, `pH =3 or - log [H^(+)] =3 or log [H^(+)] =- 3=overset(-)(3).0` ` [H^(+)] = " Antilog "[H^(+)] = " Antilog "(overset(-)(3).0) = 10^(-3)M` Similarly ,for second solution, " "`pH=4 and [H^(+)] =10^(-4) M` For third solution " "`pH =5 and [H^(+)] =10^(-5) M` Total `[H^(+)]` in the solution `=(10^(-3)+ 10^(-4) +10^(-5))=10^(-3) (1+10^(-1) +10^(-2))` `=10^(-3)(1+0.1 +0.01) = 1.11 xx 10^(-3)M` Since equal volumes of the three solutions have been mixed. `[H^(+)]` of the resulting solutin `=((1.11 xx 10^(-3)M))/(3) = 3.7 xx 10^(-4) M` |
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| 175. |
Why gas fizzes out when soda water bottle is opened ? |
| Answer» The amount of he gas dissolved is very high pressure. On opening the bottle , the pressure tends to decrease to atmospheric pressure. So the solubility decreases , i.e., the dissolved gas escapes out. | |
| 176. |
The standard Gibbs energy change for a gaseous reaction at 27 is X Kcal. If equilibrium constant a reaction is 100 and R is 2 cal K−1 mol−1. Then X is: |
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Answer» ΔGo=−RT lnK ∴X = −2 × 300 × ln 100 ∴ X = −2.7636 K cal |
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| 177. |
The solubility product of a sparingly soluble metal hydroxide `M (OH)_(2)` at 298 K is `5xx10^(-16) "mol"^(3) "dm"^(-9)`. The pH value of its queous and saturated solution isA. 5B. 9C. `11.5`D. `2.5` |
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Answer» Correct Answer - B `{:(M(OH)_(2),hArr,M^(2+),+,2OH^(-)),(,,s,,2s):}` `K_(sp)=(s) (2s)^(2)=4s^(3)` `s^(3)=(K_(sp))/(4)=(5xx10^(-16))/(4)=1.25xx10^(-16)` `=125xx10^(-18)` `:. S=5xx10^(-6)M, i.e., [OH^(-)]2xx5xx10^(-6)M` `=10^(-5)M` pOH = 5.0 pH=14-5 = 9 |
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| 178. |
The conjugate acid of amide ion `(NH_(2)^(-))` isA. `N_(2)H_(4)`B. `NH_(2)OH`C. `NH_(4)^(+)`D. `NH_(3)` |
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Answer» Correct Answer - D Whenever a species accepts a proton, it forms a conjugate acid. Thus, to find the conjugated acid of amide ion `(NH_(2)^(-))`, we add `H^(+)` and put one more positive charge (i.e., increase charge by 1), `H^(+)+underset("Base")(NH_(2)^(+))rarr underset("Acid")(NH_(3))` |
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| 179. |
(a) Calculate pH of 1.0 × 10–8 M solution of HCl.(b) The species: H2O, HSO4– and NH3 can act both as Bronsted acids and bases. For each case, give the corresponding conjugate acid and conjugate base. |
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Answer» (a) H2O(l) + H2O(l) ⇌ H2O+ (aq) + OH– (aq) Kw = [H3O+ ] [OH–] = 10–14 Let x = [OH–] = [H3O+ ] from H2O The H3O+ concentration is generated (i) from the ionization of HCl dissolved i.e. HCl(aq) + H2O(l) ⇌ H3O + (aq) + Cl– (aq) and (ii) from ionization of H2O. In these very dilute solutions, both sources of H3O+ must be considered: [H3O+] = 10–8 + x \(\therefore\) Kw = (10–8 + x) (x) = 10–14 or x2 + 10–8 x – 10–14 = 0 The value of x can be determined from quadratic equation x = 9.5 × 10–8 [OH– ] = 9.5 × 10–8 Now, pOH = – log [OH– ] = – log 9.5 + 8 log 10 = – 0.098 + 8 pOH = 7.02 \(\because\) pH + pOH = 14 pH + 7.02 = 14 or pH = 14 – 7.02 = 6.98 Hence, pH of 10–8 M solution of HCl is 6.98. (b)
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| 180. |
The pH of 0.1 M monobasic acid is 4.50. Calculate the concentration of species H+ , A– and HA at equilibrium. Also, determine the value of Ka and pKa of the monobasic acid. |
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Answer» Given C = 0.1 M pH = 4.50 \(\therefore\) pH = – log [H+] or [H+] = 10–pH = 10–4.50 = 3.16 × 10–5 [H+] = [A–] = 3.16 × 10–5 For the reaction, HA ⇌ H+ + A– Ka = \(\frac{[H^+][A^-]}{[HA]}\) = \(\frac{(3.16\times10^{-15})(3.16\times10^{-5})}{0.1}\) [∵ [HA]equi = 0.1 – (3.16 × 10–5 ) ~ 0.1] = 1.0 × 10–8 \(\because\) pKa = – log [Ka] = – log (1 × 10–8 ) = 8 log 10 = 8 |
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| 181. |
Calculate the pH of 0.08 M solution of hypochlorous acid, HOCl. The ionization constant of the acid is `2.5xx10^(-5)`. Determine the percent dissociation of HOCl. |
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Answer» Correct Answer - `2.85, 1.76%` `{:(,HOCl,+,H_(2)O,hArr,H_(3)O^(+),+,ClO^(-)),("Initial conc.",0.08,,,,,,),("At. Eqm.",0.08-x,,,,x,,x" (x=Amount of HOCl dissociated)"),(,~= 0.08,,,,,,):}` `K_(a)=(x^(2))/(0.08) or x=sqrt((2.5xx10^(-5))(0.08))=1.41xx10^(-3)` `pH = - log [H_(3)O^(+)]=-log (1.41xx10^(-3))=2.85` Alternatively, directly `pH = (1)/(2) [pK_(a) - log C]` % dissociation = `("Amount dissociated")/("Amount taken")xx100=(1.41xx10^(-3))/(0.08) xx100=1.76%` |
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| 182. |
Calculate the pH of 0.15 M solution of hypochlorous acid HClO `(K_(a)=9.6xx10^(-7))`. |
| Answer» Correct Answer - 3.42 | |
| 183. |
Calculate the pH of a solution that contains 1.00 M HF `(K_(a)=7.2xx10^(-4))` and 5.00 M HClO `(K_(a)=3.5xx10^(-8))`. |
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Answer» As `K_(a)` for `HClO lt lt K_(a)` for HF, `[H_(3)O^(+)]` obtained from HClO can be neglected in comparison to `[H_(3)O^(+)]` obtained from HF. Hence `[H_(3)O^(+)]=sqrt(K_(a) (HF)xxC(HF))=sqrt((7.2xx10^(-4))(1.0))=2.68xx10^(-2)M` `pH = - log (2.68 xx 10^(-2))=2-0.4265 = 1.57` (approx) |
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| 184. |
Mention the factors that affect equilibrium constant. |
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Answer» Temperature, pressure, catalyst and molar concentration of reactants and products. |
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| 185. |
Name the factors that affect equilibrium position of a reversible reaction. |
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Answer» (i) Temperature (ii) Pressure, (iii) Concentration, (iv) Catalyst. |
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| 186. |
Write the reversible reaction taking place between ferric ions and thiocyanate ions and write the colour of each reactant and product. |
| Answer» `{:(Fe^(3)(aq),+,SCN^(-)(aq),hArr,[Fe(SCN)]^(2+)),("(Yellow)",,"(Colourless)",,"(Reddish brown)"):}` | |
| 187. |
The solubility product of `BaCl_(2)` is `3.2xx10^(-9)`. What will be solubility in mol `L^(-1)`A. `4xx10^(-3)`B. `3.2xx10^(-9)`C. `1xx10^(-3)`D. `1xx10^(-9)` |
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Answer» Correct Answer - C `BaCl_(2)hArrBa^(2+)+2Cl^(-)` `K_(sp)=[Ba^(2+)][Cl^(-)]^(2)=x xx(2x)^(2)=4x^(3)` `4x^(3)=3.2xx10^(-9)` `rArr" "x=9.28xx10^(-4)=0.928xx10^(-3)~~1xx10^(-3)` |
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| 188. |
Mark the appropriate choice to fill up the blanks in the given paragraph. A solution which maintains constant pH when small amounts of acid or base are added is known as a `ul((i))`. A mixture of acetic acid and sodium acetate acts as `ul((ii))` with a pH around `ul((iii))` and a mixture of ammonium chloride and ammonium hydroxide acts as `ul((iv))` with a pH around `ul((v))`A. `{:((i),(ii),(iii),(iv),(v),),("buffer capacity","basic buffer",9.25,"acidic buffer",4.75,):}`B. `{:((i),(ii),(iii),(iv),(v),),("buffer solution","acidic buffer",9.25,"basic buffer",4.75,):}`C. `{:((i),(ii),(iii),(iv),(v),),("buffer solution","basic buffer",4.75,"acidic buffer",9.25,):}`D. `{:((i),(ii),(iii),(iv),(v),),("buffer solution","acidic buffer",4.75,"basic buffer",9.25,):}` |
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Answer» Correct Answer - D Acidic buffer has a pH of around 4.75 and basic buffer has a pH around 9.25. |
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| 189. |
In a reaction `A_(2)(g)+4B_(2)(g) hArr 2AB_(4)(g), DeltaH lt 0`. The formation of `AB_(4)` is not favoured byA. low temperature and high pressureB. high temperature nad low pressureC. low temperature and low pressrueD. high temperature and high pressrue |
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Answer» Correct Answer - A Since , the reaction is exothermic , the formation of `AB_(4)` is favoured by low temperature . As the forward reaction is accompanied by decrease in the number of moles, so if is favoured by high pressure. |
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| 190. |
STATEMENT-1: For `H_(2)O(l)hArrH_(2)O(g)` vapour pressure if P atm then `K_(rho)` is equal to vapour pressure STATEMENT-2: K`_(rho)` can be changed by adding more `H_(2)O` vapour from our sideA. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-9B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-9C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
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Answer» Correct Answer - C |
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| 191. |
How much of 0.3 M ammonium hydroxide should be mixed with 30 mL of 0.2 M solution of ammonium chloride to give a buffer solution of pH 10. Given `pK_(b)` for `NH_(4)OH` is 4.75. |
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Answer» Correct Answer - 112.5 mL `pOH = 14 - pH = 14 - 0 = 4 ` `pOH = pK_(b) + log .(["Salt"])/(["Base"]), 4= 4.75 + log.(["Salt"])/(["Base"]) or log.(["Salt"])/(["Base"]) = - 0.75 = 1.25 or (["Salt"])/(["Base"]) = 0.1778` i.e.,`("Moles of salt")/("Moles of base")=0.1778 or ((0.2)/(1000)xx30)/((0.3)/(1000)xxV)=0.1778` or `V = 112.5 ml` |
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| 192. |
Buffer index of a buffer of 0.1 M `NH_(4)OH` and 0.1 M `NH_(4) C l ` is `(pK_(b)` for `NH_(4)OH=4.74)`A. 0.116B. 0.232C. 0.058D. 0.348 |
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Answer» Correct Answer - A `pOH = pK_(b) + log. (["Salt"])/(["Base"])=4.74` Suppose 1 ml of 1 M HCl (i.e. 0.001 mol of HCl) is added. It will convert 0.001 mol of `NH_(4)OH` into `NH_(4)Cl` Now `[NH_(4)Cl]=0.1+0.001=0.101 M ` and `[NH_(4)OH]=0.1-0.001 = 0.099 M` Now, `pOH = 4.74 + log .(0.101)/(0.099)= 4.74+ 0.0086` Thus, change in pH = 0.0086 `:.` Buffer index `=(dn)/(dpH)=(0.001)/(0.0086) = 0.116` |
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| 193. |
18.4 g of `N_(2)O_(4)` is taken in a 1 L closed vessel and heated till the equilibrium is reached. `N_(2)O_(4(g))rArr2NO_(2(g))` At equilibrium it is found that 50% of `N_(2)O_(4)` is dissociated . What will be the value of equilibrium constant?A. 0.2B. 2C. 0.4D. 0.8 |
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Answer» Correct Answer - C `{:(,N_(2)O_(4(g)),hArr,2NO_(2(g))),("Intial conc.",18.4,,0),("No. of moles",(18.4)/(92)=0.2,,),("At equilibrium",0.2-0.1,,2xx0.1),(,=0.1,,=0.2):}` `K_(c)=(0.2xx0.2)/(0.1)=0.4` |
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| 194. |
For the equlibrium `H_(2)O(s)hArrH_(2)O(l)` which of the following statements is true?A. The pressrue changes do not effect the equlibriumB. More of ice melts, if pressrue on the system in increasedC. More of liquid freezes, if pressure on the system is increasedD. Less of ice melts, if pressrue on the system is increased |
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Answer» Correct Answer - B |
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| 195. |
In which volume ratio `NH_(4)Cl` and `NH_(4)OH` solutions (each 1 M ) should be mixed to get a buffer solution of pH 9.80 ? `(pK_(b) "of" NH_(4)OH=4.74)`A. `1:2.5`B. `2.5:1`C. `1:3.5`D. `3.5:1` |
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Answer» Correct Answer - C Suppose `V_(1) ` mL of 1 M `NH_(4)Cl` solution is mixed with `V_(2)` mL of 1 M `NH_(4)OH` solution . `V_(1)` mL of 1 M `NH_(4) Cl = V_(1)` millimole `V_(2)` mL of 1 M `NH_(4)OH=V_(2)` millimole `V_(2)` mL of 1 M `NH_(4) OH = V_(2)` millimole pH = 9.80. Hence, pOH = 14 - 9.80 = 4.20 Now, pOH = `pK_(b) + log .(["Salt"])/(["Base"])` `:. 4.20 = 4.74 + log .(V_(1))/(V_(2))` or `log. (V_(1))/(V_(2))= - 0.54 = bar(1) . 46 :. (V_(1))/(V_(2))=0.29 = (1)/(3.5)` |
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| 196. |
For a reaction, `2SO_(2(g))+O_(2(g))hArr2SO_(3(g))`, 1.5 moles of `SO_(2)` and 1 mole of `O_(2)` are taken in a 2 L vessel. At equilibrium the concentration of `SO_(3)` was found to be 0.35 mol `L^(-1)` The `K_(c)` for the reaction would beA. `5.1"L mol"^(-1)`B. `1.4"L mol"^(-1)`C. `0.6"L mol"^(-1)`D. `2.95"L mol"^(-1)` |
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Answer» Correct Answer - A `{:(,2SO_(2(g)),+,O_(2(g)),hArr,2SO_(3(g))),("Initial conc.",(1.5)/(2),,(1)/(2),,0),("At equilibrium ",(1.5)/(2)-0.35,,(1)/(2)-0.35,,0.35),(,=0.4,,=0.15,,=0.35):}` `K_(c)=((0.35)^(2))/(0.4xx0.4xx0.15)=5.1" L mol"^(-1)` |
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| 197. |
The equlibrium constnt `(K_(rho))` for the reaction, `2SO_(2)(g)+O_(2)(g)hArr2SO_(3)(g)` at 1000K is `3.5 atm^(-1)`then find out equlibrium constnt `(K_(rho))` for the reaction, `2SO_(3)(g)hArr2SO_(2)(g)+O_(2)(g)`A. `0.35` atmB. `3.5`atmC. `2.85` atmD. `0.285` atm |
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Answer» Correct Answer - D |
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| 198. |
The dissociation of ammonium hydrogen sulphide in a closed container produces a pressure of `10 atm` at `200^@C`. The value of `K_p` isA. `25`B. `50`C. `100`D. `75` |
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Answer» Correct Answer - A `{:(,NH_(4)^(+)HS^(-)(s)hArrNH_(3)(g)+H_(2)S(g)),("Initial moles (assume)"," a 0 0"),("Change"," -x +x +x"),("Equilibrium moles",bar(" a-x x x")):}` Total moles of gases `=x+x=2x` (`NH_4HS` is a solid. Thus, its moles are not counted. ) `P_(NH_3)=`(Mole fraction `_NH_3`) (Total pressure) `=(("Moles"_(NH_3))/("Total moles"))("Total pressure")` `=((x)/(2x))*P` Similarly, `P_(H_2S)=((x)/(2x))*P` `K_P=P_(NH_3)P_(H_2S)=((x)/(2x)*P)((x)/(2x)*P)` `=(P/2)^2` Since `P=10atm`, we have `K_P=((10)/(2))^2=(5)^2=25` Note that `NH_4HS` being a solid does not appear in `K_p` expression. |
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| 199. |
Out of following pair , point out the stronger Lewis acid and assign the reason: (a) BF3 and BH3 (b) Sn+2 and Sn+4 |
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Answer» (a) BF3 due to greater electronegativity of fluorine it withdraws more electrons from central 'B' atom (b) Sn+4 Greater is the charge on the ion greater is its Lewis acidic strength |
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| 200. |
Find the pH of a buffer solution having equal volumes of `0.2MNH_(4)OH and 0.2M NH_(4)Cl (K_(b)` for base `=1.0xx10^(-5)`) |
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Answer» Correct Answer - 9 |
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