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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.
| 201. |
`K_(a)` for a monobasic organic acid is `2xx10^(-5)` wiat is pH of 0.2 M aqueous solution of its salt formed with KOH? |
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Answer» Correct Answer - 9 |
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| 202. |
For the reaction, `2SO_(2(g))+O_(2(g))hArr2SO_(3(g))` What is `K_(c)` when the equilibrium concentration of `[SO_(2)]=0.60M,[O_(2)]=0.82Mand[SO_(3)]=1.90M` ?A. `12.229Lmol^(-1)`B. `24.5Lmol^(-1)`C. `36.0Lmol^(-1)`D. `2.67xx10^(3)Lmol^(-1)` |
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Answer» Correct Answer - A `K_(c)=([SO_(3)]^(2))/([SO_(2)]^(2)[O_(2)])=((1.9)^(2))/((0.6)^(2)(0.82))=12.229" L mol"^(-1)` |
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| 203. |
`NH_(4)COONH_(2)(s)hArr2NH_(3)(g)+CO_(2)(g)` If equilibrium pressure is 3 atm for the above reaction, then `K_(p)` for the reaction isA. `4`B. `27`C. `4//27`D. `1//27` |
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Answer» Correct Answer - A `{:(,NH_(4)COOH_(2)(s)hArr,2NH_(3)(g),+CO_(2)(g)),("Initial moles",a,0,0),("Equilibrium moles",a-x,2x,x):}` Partial pressure `= ("Moles fraction") x ("Total pressure")` `:. P_(NH_(3))= (2x)/(3x)P_(t)= (2)/(3)P_(t)` `P_(CO_(2))=(x)/(3x)P_(t)= (1)/(3)P_(t)` where `P_(t)` is the total pressure at equilibrium. (Note that we do not consider moles of `NH_(4)COONH_(2)` as it is a solid.) According to the law of chemical equilibrium, we have `K_(P)=P_(NH_(3))^(2)P_(CO_(2))` `= ((2)/(3)P_(t))^(2)((1)/(3)P_(t))` Since `P_(t)= 3` atm, we have `K_(P)= ((2)/(3).3)^(2)((1)/(3).3)` `=2^(2)` `=4` |
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| 204. |
`{:("Column I",,"Column II"),((A) CH_(3)COONa,,(i) "Almost neutral"pH gt 7 "or "lt7),((B) NH_(4)CI,,(ii) "Acidic"pHlt7),((C)NaNO_(3) ,,(iii)"Alkaline "pH gt 7),((D) CH_(3)COONH_(4),,(iv)"Neutral "pH =7):}`A. `(A)to(i),(B)to(ii),(C)to(iii),(D)to(iv)`B. `(A)to(ii),(B)to(iii),(C)to(iv),(D)to(i)`C. `(A)to(iii),(B)to(ii),(C)to(iv),(D)to(i)`D. `(A)to(iv),(B)to(i),(C)to(iii),(D)to(ii)` |
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Answer» Correct Answer - C `(A)CH_(3)COONahArrunderset("Weak acid")(CH_(3)COO^(-))+underset("Strong base")(Na^(+))` (Basic) (B) `NH_(4)ClhArrunderset("Weak base")(NH_(4)^(+))+underset("Strong acid")(Cl^(-))` (Acidic) (C) `NaNO_(3)hArrunderset("Strong base")(Na^(+))+underset("Strong acid")(NO_(3)^(-))` (Neutral) (D) `CH_(3)COONH_(4)hArrunderset("Weak acid")(CH_(3)COO^(-))+underset("Weak base")(NH_(4)^(+))` (Almost neutral) May be slightly acidic or basic depending upon `pK_(a) and pK_(b)` value of acid and base. |
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| 205. |
Arrange the following in increasing order of pH. `KNO_(3) (aq), CH_(3)COONa (aq), NH_(4)Cl (aq), C_(6)H_(5) CO ONH_(4) (aq)` |
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Answer» `KNO_(3)` (salt of strong acid-strong base), solution is neutral, pH = 7 . `CH_(3)CO ON a ` (salt of weak acid-strong base), solution is basic , `pH gt 7` `NH_(4)Cl` (salt of strong acid-weak base), solution is acidic, `pH lt 7` `C_(6)H_(5)CO ONH_(4)` (both weak but `NH_(4)OH` is slightly stronger than `C_(6)H_(5)CO OH`), pH close to 7 but slightly `gt 7`. Hence, in order of pH , `NH_(4)Cl lt C_(6)H_(5) CO ONH_(4) gt KNO_(3) ltCH_(3)CO ON a`. |
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| 206. |
Between `Na^(+) and Ag^(+)` which is stronger Lewis acid and why ? |
| Answer» `Ag^(+)`. This is because `Ag^(+)` has a pseudo boble gas configuration and hence has greater polarizing power. | |
| 207. |
`NH_(4)COONH_(2)(s)hArr2NH_(3)(g)+CO_(2)(g)` If equilibrium pressure is 3 atm for the above reaction, then `K_(p)` for the reaction is |
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Answer» Correct Answer - 4 |
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| 208. |
Arrange the following in increasing order of ph: `KNO_(3)(aq)`,`CH_(3)COONa(aq)`,`NH_(4)CI(aq)`,`C_(6)H_(5)COONH_(4)(aq)` |
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Answer» Increasing order of ph implies decreasing acidic strength and increasing basic strength. The correct order of increasing ph is `NH_(4)CI(aq),C_(6)H_(5)COONH_(4)(aq),KNO_(3)(aq),CH_(3)COONa(aq)` |
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| 209. |
`NH_(4)COONH_(2)(s)hArr2NH_(3)(g)+CO_(2)(g)` If equilibrium pressure is 3 atm for the above reaction, then `K_(p)` for the reaction isA. 4B. 27C. `4/27`D. `1/27` |
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Answer» Correct Answer - A |
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| 210. |
What is `K_(c )` for the following equilibrium concentration of each substance is: `[SO_(2)]=0.60 M, [O_(2)]=0.82 M` and `[SO_(3)]=1.90 M`? `2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g)` |
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Answer» `2SO_(2)(g) +O_(2)(g) hArr 2SO_(3)(g)` `K_(c) =[[SO_(3)]^(2)]/[[SO_(2)]^(2)[O_(2)]]=((1.9M)xx(1.9m))/((0.6m)xx(0.6 M)xx(0.82 M))` `=12.229 M^(-1) =12 .229 L mol^(-1)` |
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| 211. |
Aspirin (acetyl salicyclic acid, molar mass `180 g "mol"^(-1)`) used as analgesic has `pK_(a)` value of 2. Two tablets of aspirin each weighing 90 mg are dissolved in 100 mL of water. The pH of the solution isA. 0.5B. `1.0`C. `2.0`D. `4.0` |
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Answer» Correct Answer - C Aspirin is a weak acid. Hence, its pH is given by `pH=(1)/(2)[pK_(a)-log c]` Here, `c=2xx0.09 g "in " 100 mL = 1.8 g L^(-1)` `=(1.8)/(180)M=0.01 M` `:. pH=(1)/(2) [2-log 0.01]=(1)/(2)[2-(-2)]=2` |
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| 212. |
The `pK_(a) ` of acetic acid and `pK_(b)` of ammonium hydroxide are 4.76 and 4.75 respectively. Calculate the hydrolysis constant of ammonium acetate at 298K and also the degree of hydrolysis and pH of (a) 0.01 M and (b) 0.04 M solutions. |
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Answer» Correct Answer - `K_(h)=3.25xx10^(-5), h=5.7 xx 10^(-3) and pH = 7. 005` (same in both cases) First calculate `K_(a) and K_(b)`. Then `K_(h)=K_(w)//(K_(a).K_(b)) and h = sqrt(K_(h))`. |
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| 213. |
Statement-1 An aqueous solution of `CH_(3)COONH_(4)` can act as buffer. Statement -2 An aqueous solution of pure salt can act as buffer.A. Statements -1 is true , statement -2 is also true, statement -2 is the correct explanation of statement -2B. Statement -1 is true , statement-2 is also true, statement-2 is not the correct explanation of statement-2C. Statement -1 is true, statement -2 is false.D. Statement -1 is false, satement-2 is true. |
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Answer» Correct Answer - C correct statement-2 An aqueous solution of salt of weak acid and weak base can act as buffer. |
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| 214. |
At `90^(@)C` , pure water has `[H_(3)O^(+)]` as `10^(-6)` mol `L^(-1)`. What is the value of `K_(w)` at `90^(@)C` ?A. `10^(-14)`B. `10^(-6)`C. `10^(-12)`D. `10^(-8)` |
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Answer» Correct Answer - C `K_(w)=C_(H_(3O^(+))C_(OH^(-))` for pure water (neutral), `C_(H_(3)O^(+))=C_(OH^(-))`. Thus, `K_(w)=(10^(-6))(10^(-6))=10^(-12)` |
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| 215. |
If `pK_(a) `of acetic acid and `pK_(b)` of ammonium hydroxide are 4.76 each, the pH of ammonium acetate isA. 7B. less than 7C. more than 7D. zero |
| Answer» Correct Answer - A::B::C::D | |
| 216. |
Assertion. An aqueous solution of ammonium acetate can act as buffer. Reason. An aqueous solution of any pure salt acts as a buffer.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 - C Correct R. An aqueous solution of a salt of weak acid and weak base acts as buffer. |
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| 217. |
Nicotinic acid `(K_(a)=1.4xx10^(-5)) ` is represented by the formula HNiC. Calculate its percentage dissociation in a solution which contains 0.10 mole of nicotinic acid per 2.0 litre of solution. |
| Answer» Correct Answer - `1.67%` | |
| 218. |
The `pK_(a)` of acetic acid is 4.74 . The concentration of `CH_(3)C O O H ` is 0.01 M. The pH of `CH_(3) C O O H` isA. 3.37B. 4.37C. 4.74D. 0.474 |
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Answer» Correct Answer - A For a weak acid, `pH=(1)/(2) [pK_(a)-log c]` `=(1)/(2)[4.74-log 10^(-2)]=(1)/(2)xx6.74=3.37` |
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| 219. |
Asseration : An aqeous solution of ammonium acetate can act as buffer. Reason: Acetic acid is a weak acid and `NH_(4)OH` is a weak base.A. Both A and R are true and R is correct explanation of A.B. Both A and R are true but R is not correct explanation of A.C. A is false but R is true.D. Both A and R are false. |
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Answer» Correct Answer - C Correct assertion : An aqueous solution of ammonium acetate of its own cannot act as buffer. |
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| 220. |
Consider the reaction where `K_(p) = 0*497 "at 500 K`` PCl_(5) (g) hArr PCl_(3) (g) + Cl_(2) (g) ` If the three gases are mixed in a rigid container so that the partial pressure of each gas is initially 1 atm, which is true ?A. More `PCl_(5)` will be producedB. More `PCl_(3)` will be producedC. Equilibrium will be established when 50% of the reaction is completeD. None of the above |
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Answer» Correct Answer - A `Q_(p) = (p_(PCl_(3))xxp_(Cl_(2)))/(p_(PCl_(5)))=(1 atm xx1 atm )/(1atm )=1` As `Q_(p) gt K_(c)` , equilibrium will go in the backward direction , i.e., more `PCl_(5)` will be produced . |
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| 221. |
Nicotinic acid `(K_(a)=1.4xx10^(-5))` is represented by the formula HNiC. Calculate its percentage dissociation in a solution which contains 0.10 mole of nicotinic acid per 2.0 litre of solution. |
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Answer» Suppose degree of dissociation of nicotinic acid = `alpha` `{:(,HNiC,hArr,H^(+),+,NiC^(-)),("Initial amount",0.1 ",mole",,,,),("Amount",0.1-0.1 alpha,,0.1 alpha,,0.1 alpha),("at eqm.",=0.1 (1-alpha),,,,),("Molar conc.",=0.1(1-alpha)//2,,0.1 alpha//2,,0.1 alpha//2),("at eqm.",,,,,):}` `K_(a)=([H^(+)][NiC^(-)])/([HNiC])` or `1.4xx10^(-5)=((0.1 alpha)^(2)(0.1alpha//2))/(0.1(1-alpha)//2)` If `alpha lt lt 1`, then `1.4xx10^(-5)=((0.05 alpha)^(2))/(0.05)=0.05 alpha^(2)` or `alpha^(2)=2.8xx10^(-4)` or `alpha=1.67xx10^(-2)` % dissociation `= 1.67xx106(-2)xx100=1.67%` |
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| 222. |
Calculate the pH of 0.05 M sodium acetate solution of the `pK_(a)` of acetic acid is 4.74. |
| Answer» Correct Answer - 8.72 | |
| 223. |
Assertion. Aqueous solution of ammonium carbonate is basic. Reason. Acidic or basic nature of a salt solution depends upon `K_(a) and K_(b) ` value of acid and base forming the salt.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 - A R is the correct explanation of A. |
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| 224. |
Asseration : A queous solution of ammonium carbonate is basic. Reason : Acidic/basic nature of a salt of weak acid base depends on `K_(a)` and `K_(b)` value of the acid and the base forming it.A. Both A and R are ture and R is correct explanation of A.B. Both A and R are true but R is not correct explanation of A.C. A is the false but R is true.D. Both A and R are false. |
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Answer» Correct Answer - A Reason is the correct explanation for assertion. |
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| 225. |
The ionisation of hydrochloric in water is given below:HCl(aq) + H2O (l ) ⇄ H3O+ (aq) + Cl– (aq)Label two conjugate acid-base pairs in this ionisation. |
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Answer» HCl Cl– acid conjugate base H2O H3O+ base conjugate acid |
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| 226. |
What happence to the equilibrium ` PCl_(5) (g) hArr PCl_(3) (g) + Cl_(2) (g),` if nitrogen gas is added to it (i)at constant volume(ii) at constant pressure ? Give reasons. |
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Answer» (i) The state of equilibrium remains unaffected (ii) Dissociation increases (i.e., equilibrium shifts forward ).For reason |
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| 227. |
The `K_(p)` value for the reaction. `H_(2) +I_(2) hArr 2Hi` at `460^(@)C` is 49. If the initial pressure of `H_(2) " and " I_(2)` is 0.5 atm respectively , what will be the partial pressure of `H_(2)` at equilibrium ?A. `0.111` atmB. `0.123` atmC. `0.133` atmD. `0.222` atm |
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Answer» Correct Answer - A `{:(,H_(2) (g) ,+, I_(2) ,hArr, 2HI(g)),("Initial (atm)",0.5,,0.5,,0),("pressure",,,,,),("Final (atm)",0.5-x,,0.5-x,,2X),("pressure",,,,,):}` `K_(p) =((pHI)^(2))/((pH_(2))(pI_(2)))` `49 = ((2X)^(2))/((0.5-X)^(2)) " or " 7 =(2X)/((0.5 -X))` `3.5 xx -7X =2X " or " X =(3.5)/(9) =0.389` `pH_(2)` at eqm. =(0.5 - 0.389) = 0.111 |
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| 228. |
The following equilibrium exists in a closed vessel in `1L` capacity `A(g)+3B(g)hArr4C(g)` initial cocentration of `A(g)` is equal to that `B(g)`. The equilibrium concentration of `A(g)` and `C(g)` are equal. `K_(c)` for the reaction isA. `0.08`B. `0.8`C. 8D. 80 |
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Answer» Correct Answer - C |
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| 229. |
If pressure is increased on the equlibrium `N_(2)+O_(2)hArr2NO` the equlibrium willA. Shift in the forward directionB. Shift in the backward directionC. Remain undistrubedD. None of these |
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Answer» Correct Answer - C |
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| 230. |
In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.Assertion (A) : A solution containing a mixture of acetic acid and sodium acetate maintains a constant value of pH on addition of small amounts of acid or alkali.Reason (R): A solution containing a mixture of acetic acid and sodium acetate acts as a buffer solution around pH 4.75.(i) Both A and R are true and R is correct explanation of A.(ii) Both A and R are true but R is not the correct explanation of A.(iii) A is true but R is false.(iv) Both A and R are false. |
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Answer» (i) Both A and R are true and R is correct explanation of A. |
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| 231. |
One mole of ammonia was completely absorbed in one litre solution each of `(1) 1 M HCI , (2) 1 M CH_(3)COOH " and "(3) 1 M H_(2)SO_(4)` at 298 K. The decreasing order for the pH of the resulting solution is (Given , `K_(b) (NH_(3)) = 4.74 )`A. `2 gt 3 gt 1`B. `1 gt 2 gt 3`C. `2 gt 1 gt 3`D. `3 gt 2 gt1` |
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Answer» Correct Answer - C Increasing order of acidic strength after absorbing `NH_(3)` `underset((3))(H_(2)SO_(4))" "underset((1))(HCI)" "underset((2))(CH_(3)COOH)` Decreasing order of `pH : 2 gt 1 gt 3` |
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| 232. |
At a particular temperature and atmospheric pressure, the solid and liquid phases of a pure substance can exist in equilibrium. Which of substance can exist in equilibrium. Which of the following term defines this temperature ?A. Normal melting pointB. Equilibrium temperatureC. Boiling pointD. Freezing point |
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Answer» Correct Answer - A::D Solid `hArr` Liquid equilibrium exists at normal melting point or normal freezing point. |
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| 233. |
In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.Assertion (A) : Aqueous solution of ammonium carbonate is basic.Reason (R) : Acidic/basic nature of a salt solution of a salt of weak acid and weak base depends on Ka and Kb value of the acid and the base forming it.(i) Both A and R are true and R is correct explanation of A.(ii) Both A and R are true but R is not correct explanation of A.(iii) A is true but R is false.(iv) Both A and R are false. |
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Answer» (i) Both A and R are true and R is correct explanation of A. |
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| 234. |
In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.Assertion (A): The ionisation of hydrogen sulphide in water is low in the presence of hydrochloric acid.Reason (R) : Hydrogen sulphide is a weak acid.(i) Both A and R are true and R is correct explanation of A.(ii) Both A and R are true but R is not correct explanation of A.(iii) A is true but R is false(iv) Both A and R are false |
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Answer» (ii) Both A and R are true but R is not correct explanation of A. |
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| 235. |
For the reaction, `H_(2) + I_(2)hArr 2HI, K = 47.6` . If the initial number of moles of each reactant and product is 1 mole then at equilibriumA. `[1_(2)]=[H_(2)],[l_(2)]gt[Hl]`B. `[1_(2)]lt[H_(2)],[l_(2)]=[Hl]`C. `[1_(2)]=[H_(2)],[l_(2)]lt[Hl]`D. `[1_(2)]gt[H_(2)],[l_(2)]=[Hl]` |
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Answer» Correct Answer - C |
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| 236. |
The value of the equilibrium constant for the reaction : H_(2) (g) +I_(2) (g) hArr 2HI (g)` at 720 K is 48. What is the value of the equilibrium constant for the reaction : `1//2 H_(2)(g) + 1//2I_(2)(g) hArr HI (g)` |
| Answer» Correct Answer - `6,928` | |
| 237. |
The equilibrium composition for the reaction is : `{:(PCl_(3),+,Cl_(2),hArr,PCl_(5)),(0.20,,0.10,,0.40 "moles//litre"):}` What will be the equilibrium concentration of `PCl_(5) " on adding " 0.10" mole of " Cl_(2)` at rhe same temperature ? |
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Answer» Correct Answer - `0*45 "mol "L^(-1)` `K_(c) = (0*40)/(0*20 xx 0*10)=20` `" New initial conc. of " Cl_(2) = 0*10+0*10 = 0*20 " mol "L^(-1)` `" New initial conc. of " PCl_(3) and PCl_(5) " remain the same " ` Supposing x mole of `PCl_(3)` reacts , the new equilibrium concs. will be `[PCl_(3)] = 0*20 - x , [C_(2)]= 0*20 - x and [ PCl_(5)] = 0*40 + x` Putting the values in `K_(c) = ([PCl_(5)])/([PCl_(3)] [ Cl_(2)])`. ` ((0*40 + x))/((0*20 - x)(0*20-x))=20 or (0*40 +x) = 20 (0*4 + x^(2)-0*40 x)or 20 x^(2) - 9 x + 0*40 = 0` or ` x = (-b pm sqrt(b^(2) - 4 ac))/(2a)=(9pm sqrt(81-4 xx 20 xx 0*4))/(2 xx 20)=(9 pm 70)/40` =` 0*4 or 0*05 (0*4 " is impossible because x cannot be greater than " 0*2) ` Hence, `[PCl_(5)] = 0*40 + 0*05 = 0*45 " mol " L^(-1)` |
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| 238. |
The equilibrium constant for the reaction `H_(2)(g) + I_(2) (g) hArr 2HI(g)` is 0.35 at 298 K. In the following mixture at 298 K, has equilibrium been reached ? If not state on which side of the equilibrium the system is : `(i) P_(H_2) =0.10` atm and `P_(HI) = 0.80` atm and there is solid `I_(2)` in the container. `(ii) P_(H_2)= 0.55 `atm and `P_(HI) = 0.44` atm and there is solid `i_(2)` in the container. `(III) P_(H_2) =2.5` atm and `P_(Hi) =0.15` atm and there is solid `I_(2)` in the container. |
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Answer» The equilibrium constant `(K_(p))` for the reaction is `H_(2)(g) +I_(2)(g) hArr 2HI (g)` `K_(p) = (p^(2)HI(g))/(pH_(2) (g) xx pI_(2) (g)) = 0.35` Now , in all the cases , solid iodine is present in the mixture. Therefore, the partial pressure of `I_(2) i.e. pI_(2)` is taken as 1 atm in each case. Let us calculate the value of `Q_(p)` for the different reactions. `(i) " "P_(H_2) = -0.10 atm and P_(HI) =0.80 atm` `:. " "Q_(P) = (P_(HI)^(2) (g))/(P_(H_2)(g))=(0.44xx0.44)/(0.55) =0.35` As the value of `Q_(p)` (6.4) in more than that of `K_(p)` (0.35) this means that the equilibrium has not been achieved in the reaction and the reaction has proceeded more in the forward direction. `(II)" "P_(H_2) =0.55 atm and P_(HI) =0.44 atm` `:. " "Q_(p) = (P_(HI)^(2)(g))/(P_(H_2)(g)) = (0.8xx 0.8)/(0.1) = 6.4` As the value of `Q_(p)` is the same as that of `K_(p)` this means that the system is in a state of equilibrium `(iiI) " "P_(H_2) = 2.5 atm and P_(HI) = 0.15 atm` `:." "Q_(p) = (P_(HI)^(2)(g))/(P_(H_2) (g)) = (0.15xx0.15)/(2.5) = 0.009` As the value of `Q_(p)` (0.009) is less than that of `K_(p)` (0.35) , this means that the equilibrium has not been achived in the reaction and the reaction has proceeded more in the backward direction. |
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| 239. |
In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.Assertion(A): For any chemical reaction at a particular temperature, the equilibrium constant is fixed and is a characteristic property.Reason (R) : Equilibrium constant is independent of temperature.(i) Both A and R are true and R is correct explanation of A.(ii) Both A and R are true but R is not correct explanation of A.(iii) A is true but R is false.(iv) Both A and R are false. |
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Answer» (iii) A is true but R is false. |
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| 240. |
A 0.01 M solution of acetic acid is 1.34 % ionised (degree of dissociation = 0.0134) at 298 K. What is the ionization constant of acetic acid ? |
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Answer» Correct Answer - `1.80xx10^(-6)` `{:(,CH_(3)CO OH,hArr,CH_(3)CO O^(-) , +,H^(+),,),("Initial conc.",C "mol" L^(-1),,0,,0,,),("At. eqm.",c(1-alpha),, C alpha,,C alpha,,):}` `K=(C alpha.C alpha)/(C (1-alpha))=(C alpha^(2))/(1-alpha)=(0.01xx(0.0134)^(2))/(1-00.01)=1.8xx10^(-6)` |
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| 241. |
The value of the equilibrium constant for the reaction : `H_(2) (g) +I_(2) (g) hArr 2HI (g)` at 720 K is 48. What is the value of the equilibrium constant for the reaction : `1//2 H_(2)(g) + 1//2I_(2)(g) hArr HI (g)` |
| Answer» Correct Answer - `6.9` | |
| 242. |
` 0*1` mole of `PCl_(5)` is vaporised in a litre vessel at `260^(@)C`. Calculate the concentration of `Cl_(2)` at equilibrium, if the equilibrium constant for the dissociation of `PCl_(5)` is 0.0414. |
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Answer» `{:(,PCl_(5),hArr,PCl_(3),+,Cl_(2)),(" Intial conc. ",0.1 "mole",,0,,0),(" Conc. at eqm. (moles/litres)",(0.1 -x),,x,,x):}` Applying the law of chemical equilibrium, we get `K_(c) = ([PCl_(3)][Cl_(2)])/([PCl_(5)]` Here `K_(c) = 0.0414 " (Given) "` `:. 0.0414 = (x xxx)/((0.1-x)) or x^(2)/(0.1 - x) = 0.0414 or x^(2) + 0.0414 x - 0.00414 = 0` `(x= (-0.0414 pm sqrt((0.0414)^(2) - 4 xx 1 xx (-0.00414)))/2 " "[ "Using the formula " x = (-b pm sqrt(b^(2) - 4ac))/(2a)]` ( The negative value of x is meaningless and hence is rejected ) Thus, the concentration of `Cl_(2) " at equilibrium will be " 0.0468 mol L^(-1)` |
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| 243. |
Assertion (A) : For any chemical reaction at a particular temperature, the equilibrium constant is fixed and is a characteristic property. Reason (R) : Equilibrium constant is independent of temperature.A. Both A and R are true and R is the correct explanation of A.B. Both A and R are true but R is not the correct explanation of A.C. A is true but R is false.D. Both A and R are false. |
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Answer» Correct Answer - C Correct R. Equilibrium constant of a reaction depends upon temperature. |
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| 244. |
Asseration : For any chemical reaction at particular temperautre, the equilibrium constant is fixed and is a characteristic property. Equilibrium constant is independent of temperature.A. Both A and R are true and R is correct explanation of A.B. Both A and R are true but R is not correct explanation of A.C. A is true but R is false.D. Both A and R are false. |
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Answer» Correct Answer - C Correct assertion : Equilibrium constant depneds upon temperautre. |
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| 245. |
The concentration of hydrogen ion in a sample of soft drink is `3.8 xx 10^(-3)M`. What is its `pH`?A. `3.84`B. `2.42`C. `4.44`D. `1.42` |
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Answer» Correct Answer - B `pH =- log [H^(+)] =- log (3.8 xx 10^(-3)]` `=- [log (3.8) +log (10^(-3)]` `=-[(0.58)+(-3.0)]=+2.42` |
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| 246. |
In the reaction at constant volume `C_((s)) +CO_(2) (g) hArr 2CO_((g))` argon gas is added which does not takes part in the reaction. Choose the correct statement.A. the equilibrium constant is unchangedB. The equilibrium shifts in the forward directionC. The equilibrium shifts in the backward directionD. The direction of equilibrium depends on the amount of argon added. |
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Answer» Correct Answer - A The equilibrium constant remains uchanged. |
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| 247. |
Which of the following is not the characteristic of chemical equilibrium ?A. At equilibrium, the concentration of each of the reactants and the products becomes constant.B. At equilibrium, the rate of forward reaction becomes equal to the rate of backward reaction, and hence, the equilibrium is dynamic in mature.C. A chemical equilibrium can be estabilised only if none of the products is allowed to escape out separate out as a solid.D. Chemical equilibrium for the reversible reaction `N_(2)+3H_(2)hArr2NH_(3)` can be attained from forward direction only. |
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Answer» Correct Answer - D Chemical equilibrium can be attained from either direction, i.e., from the direction of the reactants taking `N_(2)` and `H_(2)` and getting `NH_(3)` as well as from the direction of the products (taking `NH_(3)` and decomposing it into `N_(2)` and `H_(2)`). |
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| 248. |
Phosphorous pentachloride dissociates as follows (in a closed reaction vessel): `PCI_(5)(g)hArrPCI_(3)(g)+CI_(2)(g)` If the total pressure at equilibrium on the reaction mixture is P and the degree of dissociation of `PCI_(5)` is x, the partial pressure of `PCI_(3)` will beA. `((x)/(x+1))P`B. `((x)/(1-x))P`C. `((2x)/(1-x))P`D. `((x)/(x-1))P` |
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Answer» Correct Answer - A `{:(,PCl_(5)(g),hArrPCl_(3)(g),+Cl_(2)(g)),("Initial (moles)",I,0,0),("Change (moles)",1-x,+x,+x),("Equilibrium (moles)",1-x,x,x):}` Total moles `=n_(PCl_(5))+n_(PCl_(3))+n_(Cl_(2))` `=(1-x)+(x)+(x)` `=1+x` Partial pressure of `PCl_(3)` `=("Mole fraction of" PCl_(3))xx("Total pressure of" PCl_(3))` `=((x)/(1+x))P` |
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| 249. |
`K_c` and `K_p` for heterogeneous equilibrium: Write the equilibrium constant expressions `K_c` and `K_p`, if applicable, for each of the following heterogeneous system: (i) `S(s)+H_2SO_3(aq.)hArr H_2SO_3(aq.)` (ii) `2NH_3(g)+H_2SO_4(l)hArr (NH_4)_2SO_4(s)` (iii) `P_4(s)+6Cl_2(g) hArr 4PCl_3(l)` (iv) `AgCl(s)hArr Ag^(+)(aq.)+Cl(aq.)` (v) `(NH_4)Se(s) hArr 2NH_3(g)+H_2Se(g)` |
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Answer» Strategy: Apply the definitions of `K_c` and `K_p` to each reaction. Remember that in heterogeneous equilibrium, pure solids or pure liquids must be present (however, small it may be) for the equilibrium to exist, but their active masses do not appear in the expression of the equilibrium constant because they are in incoroprated into the value of equilibrium constant. Solution: (i) `K_c=(C_(H_2S_2O_3))/(C_(H_2SO_3))`, `K_p` cannot be defined as no gases are involved (ii) `K_c=(1)/(C_(NH_3))`, `K_p=(1)/(P_(NH_3))` (iii) `K_c=(1)/(C_(Cl_2)^6)`, `K_p=(1)/(P_(Cl_2)^6)` (iv) `K_c=C_(Ag^(+))C_(C1^(-))`, `K_p` undefined, no gases involved (v) `K_c=C_(NH_3)^2C_(H_2Se)`, `K_p=P_(NH_3)^1P_(H_2)Se` |
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| 250. |
Write the relation between `K_(p) " and "K_(c)` for the following reactions : `"(i)"" " PCI_(5) (g) hArr PCI_(3)(g) +CI_(2) (g)` `"(ii)"" "N_(2)(g) +3H_(2)(g) hArr 2NH_(3)(g)` `"(iii)"" "2H_(2)O(g) hArr 2H_(2)(g) +O_(2)(g)` `"(iv)"" "H_(2)(g) + I_(2)(g) hArr 2HI(g)` |
| Answer» Correct Answer - `(i) K_(p) =K_(c) (RT)^(1) (ii) k_(p) =K_(c)(RT)^(-2) (iii) K_(p) =K_(c) (RT)^(0) (iv) K_(p) =K_(c)` | |