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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.
| 1001. |
At `473 K`, equilibrium constant `K_(c )` for decomposition of phosphorus pentachloride, `PCl_(5)` is `8.3xx10^(-3)`. If decomposition is depicted as, `PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g) Delta_(r)H^(Θ)=124.0 kJ mol^(-1)` a. Write an expression for `K_(c )` for the reaction. b. What is the value of `K_(c )` for the reverse reaction at the same temperature? c. What would be the effect on `K_(c )` if i. More `PCl_(5)` is added ii. Pressure is increased iii. The temperature is increased? |
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Answer» (a) The expression for `K_(c) = [[PCI_(3)(g)][CI_(2)(g)]]/[[PCI_(5)(g)]]` (b) For reverse reaction `(K_(c)) = (PCI_(5)(g))/[[PCI_(3)(g)][CI_(2)(g)]] =(1)/(8.3 xx 10^(-3)) =120.48` (c) (i) By adding more of `PCI_(5)` value of `K_(c)` wil remain constant because there is no change in temperature . (ii) By increasing the temperature , the forward reaction will be favoured since it is endothermic in nature. Therefore , the value of equilibrium constant will increase. |
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| 1002. |
At 473 K, equilibrium constant, Kc for decomposition of phosphorus pentachloride, PCl5 is 8.3 x 10-3. If decomposition is depicted asPCl5(g) ⇌ PCl3(g) + Cl2(g);ΔH = 124.0 kJ mol-1Write an expression for Kc for the reaction. |
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Answer» Kc = \(\frac{[PCl_3][Cl_2]}{[PCl_5]}\) |
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| 1003. |
CO2(g) + C(s) ⇌ 2CO(s)reaction will get affected by increase of pressure? Also mention, whether change will cause the reaction to go into forward or backward direction? |
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Answer» Δn = 2 - 1 = 1 So, the position of equilibrium for this reaction will shift backwards, (or towards) (left). |
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| 1004. |
Describe the effect of removal of CH3OH on the equilibrium of the reaction.2H2(g) + CO(g) ⇌ CH3OH(g) |
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Answer» 2H2(g) + CO(g) ⇌ CH3OH(g) Removal of CH3OH favours the forward reaction. |
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| 1005. |
Describe the effect of removal of CO on the equilibrium of the reaction.2H2(g) + CO(g) ⇌ CH3OH(g) |
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Answer» 2H2(g) + CO(g) ⇌ CH3OH(g) Removal of CO favours the reverse reaction. |
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| 1006. |
For the equilibrium, `2 NOCl (g) hArr 2 N O (g) + Cl_(2) (g),` the value of the equilibrium constant, `K_(c)" is " 3.75 xx 10^(-6)" at " 1069 K." Calculate " K_(p)` for the reaction at this temperature. |
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Answer» For the given reaction , `Delta n = (2+1) - 2 = 1` `K_(p) = K_(c)(RT)^(Delta n) = (3. 75 xx 10^(-6)) ( 0.0831 xx 1069)` `= 3.33 xx 10^(-4)`. |
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| 1007. |
CaCO3(s) ⇌ CaO(s) + CO2(g)reaction will get affected by increase of pressure? Also mention, whether change will cause the reaction to go into forward or backward direction? |
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Answer» Δn = +1 The reaction proceeds with an increase in the number of moles of the gaseous components and hence in the volume of the system. Therefore, an increase in pressure will favour the backward reaction, i.e., the reaction will move towards left (reactants side). |
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| 1008. |
Decribe the effect of: a. Addition of `H_(2)` b. Addition of `CH_(3)OH` c. Removal of `CO` d. Removal of `CH_(3)OH` on the equilibrium of the reaction: `2H_(2)(g)+CO(g) hArr CH_(3)OH(g)` |
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Answer» (i) Equilibrium will be shifed in the forward direction. (ii) Equilibrium will be shifted in the backward direction. (iii) Equilibrium will be shifted in the backward direction. (iv) Equilibrium will be shifted in the forward direction. |
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| 1009. |
For the equilibrium `2NOCl(g) hArr 2NO(g)+Cl_(2)(g)` the value of the equilibrium constant, `K_(c)` is `3.75 xx 10^(-6)` at `1069 K`. Calcualate the `K_(p)` for the reaction at this temperature? |
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Answer» We know that, `K_(p) = K_(c)(RT)^(Deltan)` For the above reaction, `Deltan = (2+1)-2=1` `K_(p) = 3.75 xx 10^(-6)(0.0831 xx 1069)` `K_(P) = 0.033` |
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| 1010. |
which of the following reactions will get affected by increase in pressure ? Also mention whether the change will cause the reaction to go the right or left direction. (i) `CH_(4)(g)+2S)_(2) (g) hArr CS_(2) (g) + 2H_(2)S(g)` (ii) `CO_(2) (g) +C(s) hArr 2CO (g)` (iii) `4NH_(3)(g) +5O_(2)(g) hArr 4NO (g) + 6H_(2)O(g)` (iv) `C_(2)H_(4)(g) +H_(2)(g) hArr C_(2)H_(6) (g)` (v) `COCI_(2)(g) hArr CO(g) +CI_(2)(g)` (vi) `CaCO_(3)(g) hArr CaO(s)+CO_(2)(g)` |
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Answer» Only those reactions will be affected by increasing the pressure in which the number of moles of the gaseous reactants and products are different `(n_(p) ne n_(r))` (gaseous). With the exception of the reaction (1),all the remaining five reactions will get affected by increasing the pressure . In general. the reaction will go to the left if `n_(p) gt n_(r)` The reaction will go to the right if `n_(p) lt n_(r)` Keeping this in mind. (i) Increases in pressure will not affect equilibrium because `n_(p) =n_(r) =3.` (ii) Increase in pressure will favour backward reaction because `n_(p)(2) gt n_(r) (1)` (iii) Increase in pressure will favour backward reaction because `n_(p) (10) gt n_(r) (9)` (iv) Increase in pressure will favour forward reaction because`n_(p)(1) lt n_(r) (2)` (v) Increase in pressure will favour backward reaction because `n_(p) (2) gt n_(r)(1)` (vi) Increase in pressure will favour backward reaction because `n_(p) (1) gt n_(r) (0)` |
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| 1011. |
Find out the value of `K_(c )` for each of the following equilibrium from the value of `K_(p)`: a. `2NOCl(g) hArr 2NO(g)+Cl_(2)(g), K_(p)=1.8xx10^(-2)` at `500 K` b. `CaCO_(3)(s) hArr CaO(s)+CO_(2)(g), K_(p)=167` at `1073 K` |
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Answer» Correct Answer - (i) `4.33 xx 10^(-4)` (ii) `1.90` The relation between `K_(p)` and `K_(c)` is given as : `K_(p) = K_(c) (RT)^(Deltan)` (a) Here, `Deltan = 3-2 = 1` `R = 0.0831 "bar L mol"^(-1) K^(-1)` `T = 500 K` `K_(p) = 1.8 xx 10^(-2)` Now, `K_(p) = K_(c)(RT)^(Deltan)` `rArr 1.8 xx 10^(-2) = K_(c) (0.0831 xx 500)^(1)` `rArr K_(c) = (1.8 xx 10^(-2))/(0.0831 xx 500)` `= 4.33 xx 10^(-4)` (approximately) (b) Here, `Deltan = 2-1 = 1` `R = 0.0831 "bar L mol"^(-1) K^(-1)` `T = 1073 K` `K_(p) = 167` Now, `K_(p) = K_(c) = (RT)^(Deltan)` `rArr 167 = K_(c) (0.0831 xx 1073)^(Deltan)` `K_(c) = (167)/(0.0831 xx 1073)` `= 1.87` (approximately) |
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| 1012. |
The species: H2O,HCO3- , 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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| 1013. |
What will the conjugate acids for the following Bronsted bases? NH2- , NH3 and HCOO- |
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Answer» NH2- , NH3 and HCOO- |
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| 1014. |
Calculate the precent dissociation of ` H_(2) S (g) " if " 0*1 " mole of " H_(2) S " is kept in " 0*4 " litre vessel at 1000 K . For the reaction . " 2H_(2) S (g) hArr 2 H_(2) (g) + S_(2) (g), " the value of " k_(c) " is " 1*0 xx 10^(-6)` |
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Answer» `H_(2) S = (0*1 )/(0*4) "mol" L^(-1) = 0*25 "mol" L^(-1) ` Suppose degree of dissociation of `H_(2) S = alpha ` . Then ` {: (,2 H_(2)S ,hArr,2H_(2),+,S_(2)),(" Intial conc.",0*25 M,,,,),("Conc.at eqm.",0*25 (1-alpha),,0*25 alpha = 0* 125 alpha ,,(0*25)/2alpha=0*125alpha):}` `K_(c) = ([H_(2)]^(2) [S_(2)])/([H_(2)S]^(2)):. 10^(-6) = ((0*25 alpha)^(2) (0*125 alpha))/[ 0*25 (1-alpha)]^(2)` Neglecting `alpha` in comparison to 1 , we get ` 10^(-6) = ((0*25 alpha)^(2) ( 0*125 alpha))/(0*25 )^(2) ` ` 10^(-6)= 0*125 alpha^(3) or alpha^(3) = 8 xx 10^(-6) or alpha = 2 xx 10^(2) = 0*02 ` ` :. % " age dissociation "= 0*02 xx 100 = 2 % ` |
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| 1015. |
Assertion. Precipitation of soap is made by the addition of salt (NaCl). Reason. Presence of common ion suppresses the dissociation of weak acid.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. Salting out action of soap is based on the principle of solubility product. Common ion effect is for weak electrolytes (weak acids or weak bases). |
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| 1016. |
Calculate the hydrogen ion concentration in the following biological fluids whose `pH` are given below: a. Human muscle-fluid, `6.83` b. Human stomach fluid, `1.2` c. Human blood, `7.38` d. Human saliva, `6.4`. |
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Answer» (a)`[H^(+)]` of human muscles fluid `pH=6.83 " or " - log [H^(+)]= 6.83 " or " log [H^(+)] =- 6.83 =bar(7).17` `[H^(+)] = " Antilog " [H^(+)] = " Antilog [bar(7).17] = 1.479 xx 10^(-7) M` (b) `[H^(+)]` of human stomach fluid `pH =1.2 "or " - log [H^(+)] =1.2 " or " log [H^(+)] =- 1.2` `[H^(+)] = " Antilog " (-1.2) = " Antilog (bar(2).8) =6.309 xx 10^(-2) M` (c) `[H^(+)]` of human blood `pH =7.38 " or " - log [H^(+)] =7.38 " or " log [H^(+)] =- 7.38` `[H^(+)] = "Antilog " (-7.38) = " Antilog"(bar(8).62) =4.168 xx 10^(-8) M` (d) `[H^(+)]` of human saliva `pH =6.4 " or " - log [H^(+)] =6.4 " or " log [H^(+)] =- 6.4` `[H^(+)] = " Antilog (-6.4) = " Antilog " (bar(7).6) =3.98 xx 10^(-7) M.` |
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| 1017. |
If `K_(1)` is ionization constant of `H_(2)S(aq)hArr2H^(+)(aq)+S^(2-)(aq) and K_(2)` is that for `H_(2)S(aq)hArr H^(+)(aq)+HS^(-)(aq)`, then ionization constant of `HS^(-)(aq)hArr H^(+)(aq)+S^(2-)(aq)` will be equal to .......... . |
| Answer» Correct Answer - `K_(1)//K_(2)` | |
| 1018. |
The pH of `10^(-8)` M acid solution lies between .........and .............. . |
| Answer» Correct Answer - 6 and 7 | |
| 1019. |
A buffer solution can be prepared from a mixture of (1).Sodium acetate and acetic acid in water (2).Sodium acetate and hydrochloric acid in water (3).ammonia and ammonium chloride in water (4). Ammonia and sodium hydroxide in water. The correct answer is :A. `1,3,4`B. `2,3,4`C. `1,2,4`D. `1,3` |
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Answer» Correct Answer - D is the correct answer. |
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| 1020. |
Calculate the pH value of 0.0001 M `HNO_(3)` |
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Answer» Correct Answer - `pH =4` `HNO_(3)` is a strong acid and it dissociates completely in aqueous solution. The concentration of the ions is the same as that of the acid `HNO_(3) overset((aq))(to) underset(0.0001 M)(H_(3)O^(+)) +underset(0.0001M)(NO_(3)^(-)(aq))` `[H_(3)O^(+)] =0.0001 M =10^(-4) M` `pH =- log [H_(3)O^(+)] =-log [10^(-4)] =(-) (-4) log 10=4` |
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| 1021. |
Calculate the pH value of (a) 0.0001 M NaOH (b) 0.01 M NaOH and (c) 0.04 M NaOH solution at `25^(@)`C. |
| Answer» Correct Answer - (a) 10 (b) 12 (c) 12.60 | |
| 1022. |
What `[H_(3)O^(+)]` must be maintained in a saturated `H_(2)S` solution to precipitate `Pb^(2+)` but not `Zn^(2+)` from `[K_(sp)H_(2)S=1.1xx10^(-22) and K_(sp)ZnS=1.0xx0^(-21)]` |
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Answer» `K_(sp) ` for `ZnA = 10^(-21) and [Zn^(2+)] = 0.01 M = 10^(-2) M` But `K_(sp) [ZnS) = [ Zn^(2+) ] [S^(-2)], i.e., 10^(-21) = [10^(-2)] [S^(2-)] or [S^(2-) ] = 10^(-19) M` Thus, to prevent precipitation of `Zn^(2+)` ions, `[S^(2-)]` must be less than `10^(19)` M. Futher, `H_(2)S overset(2H_(2)O)rarr 2 H_(3)O^(+) + S^(2-)` `:. K_(sp) (H_(2)S) = [ 2 H_(3)O^(+)]^(2) [S^(2-)]` `1.1xx10^(-22)=[2H_(3)O^(+)]^(2)xx10^(-19) or [2 H_(3)O^(+)]^(2)=1.1xx10^(-3) = 11xx10^(-4)` or `[2H_(3)O^(+)] = 3.32xx10^(-2) M or [H_(3)O^(+) ] = 1.66xx10^(-2) M` |
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| 1023. |
Calculate the pH of `10^(-10)` M NaOH solution. |
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Answer» `[OH^(-)] " given by base " = 10^(-10) M` `[OH^(-)] " given by water "=10^(-7) M` `" Total "[OH^(-)] =10^(-10) M+ 10^(-7) M= 10^(-7) (10^(-3)+1) = 1.001 xx 10^(-7) M` `pOH =- log (1.001 xx 10^(-7)) = (7 - log 1.001)` `=7 -0.0004 = 6.9996 , pH =14 -6.9996 = 7.0004` |
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| 1024. |
A sample of AgCl was treated with 5.00 ml of 1.5 M `Na_(2)CO_(3)` solution to give `Ag_(2)CO_(3)`. The remaining solution contained 0.0026 g of `Cl^(-)` per litre. Calculate the solubility product of AgCl `(K_(sp) "for" Ag_(2)CO_(3)=8.2xx10^(-2))` |
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Answer» `1.5 M Na_(2)CO_(3) ` gives `[CO_(3)^(2-)]=1.5M` `K_(sp)` for `Ag_(2)CO_(3)=[Ag^(+)]^(2)[CO_(3)^(2-)]` `:. [Ag^(+)]=sqrt((K_(sp) "for" Ag_(2)CO_(3))/([CO_(3)^(2-)]))=sqrt((8.2xx10^(-12))/(1.5))=2.34xx106(-6)M` `K_(sp) "for" AgCl=[Ag^(+)][Cl^(-)]=(2.34xx10^(-6))((0.0026)/(35.5))=1.71xx10^(-10)` |
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| 1025. |
At `18^(@)C`, the solubility product of AgCI is `1.8xx10^(-10)`. In the solution, the value of `Ag^(+)` is `4xx10^(-3)` mol `L^(-1)`. The value of `[CI^(-)]` to precipitate AgCI from this solution should be greater thanA. `4.5xx10^(-8)mol L^(-1)`B. `7.2xx10^(-12)mol L^(-1)`C. `4.0xx10^(-3)mol L^(-1)`D. `4.5xx10^(-7)mol L^(-1)` |
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Answer» Correct Answer - A Equilibrium expression is `AgCl(s)hArr Ag^(+)(aq.)+Cl^(-)(aq.)` `K_(sp)=C_(Ag^(+))C_(Cl^(-))` `1.8xx10^(-10)=(4xx10^(-3))C_(CI^(-)` or `C_(CI^(-))= (1.8xx10^(-10))/(4.0xx10^(-3))` `= 0.45xx10^(-7)mol L^(-1)` `=4.5xx10^(-8)mol L^(-1)` Since precipitation takes place whenever ionic product just exceeds solubility product, `C_(CI^(-))` should be greater than `4.5xx10^(-8)mol L^(-1)` to precipitate AgCI. |
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| 1026. |
The pH of a solution of AgCI(s) with solubility product `1.6 xx 10^(-10)` in 0.1 M Nacl solution would be :A. `1.26 xx 10^(-5) M`B. `1.6 xx 10^(-9) M`C. `1.6 xx 10^(-11) M`D. Zero |
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Answer» Correct Answer - B Let S be the solubility of AgCI in moles per litre. `underset(S)(AgCI (aq)) hArr underset(S)(Ag^(+))(aq) + underset((S+0.1))(CI^(-)(aq))` (0.1 NaCI solution also provides 0.1 M `CI^(-)` ions) `Ks_(p) =[Ag^(+)] [CI^(-)]` `1.6 xx 10^(-10) =S [S +0.1] = S(0.1)` `("Because " S lt lt lt 0.1)` `S=(1.6xx 10^(-10))/(0.1) =1.6 xx 10^(9) M` |
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| 1027. |
The pH of a solution obtained by mixing 60 mL of 0.1 M NaOH solution and 40 mL of 0.15 MHCI solution is :A. 10B. 12C. 2D. 7 |
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Answer» No. of millimoles of `NaOH =60 xx 0.1 =6` No. of millimoles of `HCI =40 xx0.15 =6` Since the no. of millimoles of `NaOH" and " HCI` are equal the solution upon mixing is neutral . Its pH is 7. |
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| 1028. |
For `PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g),` write the expression of `K_(c)` |
| Answer» `K_(c)([PCl_(3)][Cl_(2)])/([PCl_(5)])` | |
| 1029. |
Water is rather neutral but becomes a strong base when HCI is dissolved in water . Explain. |
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Answer» The basic character of water is explained on the basis of Bronsted -Lowry concept. HCI is a strong acid which means that it readily loses `H^(+)` ions. All the ions are to be accepted by `H_(2)O` molecules to form `H_(3)O^(+)` ions. This means `H_(2)O` is a strong base in contact with HCI or any other strong acid. `{:(HCI,+,H_(2)O ,hArr,H_(3)O^(+) ,+,CI^(-)),(underset(("Strong"))("Acid"-I),, underset(("Strong"))("Base-I"),,"Acid-I",,"Base-I"):}` Similarly in contact with a weak acid like `CH_(3)COOH` water behaves as a weak base. |
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| 1030. |
For the reaction `PCl_(3)(g)+Cl_(2)(g)hArrPCl_(5)(g)` the value of `K_(p)` at `250^(@)C` is `0.61atm ^(-1)` The value of `K_(c)` at this temperature will beA. `15.19(mol L^(-1))`B. `26.19(mol L^(-1))`C. `35.19(mol L^(-1))`D. `52.19(mol L^(-1))` |
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Answer» Correct Answer - B |
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| 1031. |
Formaldehyde polymerises to form glucose according to the reaction ` 6 HCHO hArr C_(6) H_(12)O_(6)` ltbr gt The theoretically computed equiibrium constant for this reaction is found to be ` 6 xx 10^(22)`. If 1 M solution of glucose dissociates according to the above equilibrium , what will be the concentration of formaldehyde in the solution ? |
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Answer» ` 6 HCHO hArr C_(6) H_(12)O_(6)` As equilibrium constant for this reaction is very large, therfore for the reverse reaction involving dissociation of glucose, equilibrium constant is very very small . Hence , for the reverse reaction ` C_(6) H_(12)O_(6) hArr 6 HCHO " " (K= 1/(6 xx 10^(22)))^(1//6)` dissociation of glucose is negligible . Starting with 1 M, concentration at equilibrium at equilibrium `cong 1 M` ` K= ([HCHO]^(6))/([C_(6)H_(12)O_(6) )` ` 1/(6 xx 10^(22)) = ([HCHO]^(6))/1 or [HCHO] = (1/(6xx10^(22)))^(1//6)` `log [HCHO] = 1/6 [-log (6xx10^(22)] = 1/6 [-22* 778 ] = -3* 7963 = bar 4* 2137` ` :. [HCHO] = 1* 636 xx 10^(-4) "M"` |
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| 1032. |
Calculate pH values of (i) 0.2 M `H_(2)SO_(4)` solution (ii) 0.2M `Ca(OH)_(2)` solution. |
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Answer» (i) pH, value of 0.2M `H_(2)SO_(4)` solution `H_(2)SO_(4)` is a dibasic acid and dissociates in solution as : `H_(2)SO_(4) hArr 2H^(+) + SO_(4)^(2-)` Concentration of `[H^(+)] = 2 xx 0.2 = 0.4 M` pH of solution `=- log [H_(3)O^(+)]=-log[0.4]=- (-0.3979)= 0.3979` (ii) pH value of 0.2M `Ca(OH)_(2)` solution `Ca(OH)_(2)` is a diacid base and dissociates in solution as : `Ca(OH)_(2) hArr Ca^(2+) + 2OH^(-)` Concentration of `[OH^(-)] = 2 xx 0.2 =0.4M` `pOH=- log (OH^(-)) =- log [0.4] =- (-0.3979) =0.3979` `pH = 14 - pOH = 14 - 0.3979 = 13. 6021` |
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| 1033. |
In which of the following reaction `K_(p) gt K_(c)`A. `N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)`B. `H_(2)(g)+I_(2)(g)hArr2HI(g)`C. `PCl_93)(g)+Cl_(2)(g)hArrPCl_(2)(g)`D. `2SO_(3)(g)hArrPCl_95)(g)` |
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Answer» Correct Answer - D |
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| 1034. |
The solubility product for silver choride is `1.2xx10^(-10)` at 298 K. Calculate the solubility of silver chloride at 298 K. |
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Answer» Silver chloride dissociates according to the equation : `AgCl(s) hArr AgCl(aq) hArr AG^(+) (aq) + Cl^(-)(aq)` Let s be the solubility of AgCl in moles per litre. Consequently, the molar concentration of `Ag^(+) and Cl^(-)` will also be s each. Substituting in the expression for solubility product of AgCl, `K_(sp)=[Ag^(+)][Cl^(-)]=s xx s =s^(2)` But `K_(sp)= 1.2xx10^(-10) ` (Given) `:. s^(2)=1.2xx10^(-10) or s = sqrt(1.2xx10^(-19))=1.1xx10^(-5) "mol " L^(-1)` |
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| 1035. |
The equilibrium constants for the following reactions `N_(2) (g) + 3H_(2)(g) hArr 2NH_(3) (g) N_(2) (g) + O_(2) (g) hArr 2NO(g)` `" and " H_(2)(g) +1//2 O_(2) (g) hArr H_(2)O (Ig) " are " K_(1) ,K_(2) " and " K_(3)` respectively. The equilibrium constant (K) for the reaction`2NH_(3) (g) + 2^(1)//2) O_(2) (g) hArr 2NO(g) +3H_(2) O(I) " is "`A. `K_(2) .K_(3)^(3)//K_(1)`B. `K_(2)^(2) K_(3)//K_(1)`C. `K_(1).K_(2)//K_(3)`D. `K_(2).K_(3)//K_(1)` |
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Answer» Correct Answer - D `2NH_(3) hArr N_(2) +3H_(2) , 1//K_(1)` `N_(2) + O_(2) hArr 2NO, K_(2)` `3H_(2) +3//2 O_(2) hArr 3H_(2)O , K_(3)^(3)` |
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| 1036. |
Which of the following will not change the concentration of ammonia in the equilibrium `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g), DeltaH=-x kJ`A. Increase of pressureB. Increase of temperatureC. Decrease of volumeD. Addition of catalyst |
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Answer» Correct Answer - D |
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| 1037. |
Out of water and silver nitrate solution, in which case silver choride will dissolve more? |
| Answer» Silver chloride will dissolve more in water than in silver mitrate solution because the latter has `Ag^(+)` ions as the common ions. Therefore, the solubility of silver chloride will be suppressed in silver nitrate solution. | |
| 1038. |
The equilibrium constants for the following reactions `N_(2) (g) + 3H_(2)(g) hArr 2NH_(3) (g) N_(2) (g) + O_(2) (g) hArr 2NO(g)` `" and " H_(2)(g) +1//2 O_(2) (g) hArr H_(2)O (Ig) " are " K_(1) ,K_(2) " and " K_(3)` respectively. The equilibrium constant (K) for the reaction`2NH_(3) (g) + 2^(1)//2) O_(2) (g) hArr 2NO(g) +3H_(2) O(I) " is "`A. `K_(1)K_(2)//K_(3)`B. `K_(2)K_(3)^(3)//K_(1)`C. `K_(2)K_(3)^(2)//K_(1)`D. `K_(2)^(2)K_(3)//K_(1)` |
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Answer» Correct Answer - B Desired equilibrium equation is `2NH_(3)(g)+(5)/(2)O_(2)(g)hArr 2NO(g)+3H_(2)O(l)` It can be obtained by reversing the first equation, multiplying the third equation by 3, and adding the second equation to them: `2NH_(3)hArr N_(2)+3H_(2), 1//K_(1)` `3H_(2)+(3)/(2)O_(2)hArr 3H_(2)O, K_(3)^(3)` `(N_(2)+O_(2)hArrr 2NO, K_(2))/(2NH_(3)+(5)/(2)O_(2)hArr 2NO+3H_(2)O, K_(eq)=(K_(2)K_(3)^(3))/(K_(1)))` |
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| 1039. |
For the reversible reaction , ` N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g) " at " 500^(@)C , " the value of " K_(p) " is " 1*44 xx 10^(-5) ` when partial pressure is measured in atmospheres. The corresponding value of `K_(c)` , with concentration in mole` litre^(-1)`, isA. ` 1*44 xx 10^(-5) //(0* 082 xx 500) ^(-2)`B. ` 1*44 xx 10^(-5) //(8*314 xx 773 )^(-2)`C. ` 1*44 xx 10^(-5) //(0*082 xx 773)^(2)`D. ` 1*44 xx 10^(-5) // (0*082 xx 7773 )^(-2)` |
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Answer» Correct Answer - D ` K_(c) = K_(p) //(RT)^(Delta n) = 1*44 xx 10^(-5) //(0*082 xx 773 )^(-2)` |
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| 1040. |
At `500^(@)C` the equilibrium constant for the reaction `N_(2)(g) + 3H_(2) (g) hArr 2NH_(3)(g) is 6.02 xx 10^(-2) litre^(-2) mol^(-2)` What is the value of `K_(p)` at the same temperature? |
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Answer» According to the available data : `K_(c) = 6.02 xx 10^(-2) litre^(-2) mol^(-2), Delta ng = 2-4 = -2.` `R=0.082 " litre atm " K^(-1) mol^(-1), T=500^(@)C = 773 K ` By using the relation ` K_(p) = K_(c) (RT) Delta ng ` `= (6.02 xx 10^(-2) litre^(-2) mol^(-2)) xx (0.082 " litre atm " K^(-1) mol^(-1) xx 773 K)^(-2)` ` = 1.5 xx 10^(-5) atm^(-2)` |
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| 1041. |
At 773 K, the equilibrium constant `K_(c)` for the reaction, `N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g)" is " 6.02 xx 10^(-2)L^(2) mol^(-2).` Calculate the value of `K_(p)` at the same temperature. |
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Answer» `Delta n_(g) = 2 - 4 = -2, K_p= K_(c) (RT)^(Delta n)` `=6.02 xx 10^(-2) L^(2) mol^(-2) xx (0*0821 L " atm " K^(-1)"mol"^(-1) xx 773 K)^(-2)` ` = 1.5 xx 10^(-5) "atm"^(-2)` |
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| 1042. |
If concentration are expressed in moles `L^(-1)` and pressure in atmospheres, what is the ratio of `K_(p) " to " K_(c)` for the reaction : ` 2 SO_(2) (g) + O_(2) (g) hArr 2 SO_(3) (g) " at " 25^(@) C`? |
| Answer» ` Delta _(n_(g)) = n_(p) - n_(r) = -1. "Hence ", K_(p) = K_(c) (RT)^(-1) or K_(p)//K_(c)= 1//RT= 1/(0*0821 xx 298) = 0*04` | |
| 1043. |
For the reactions `N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g), " at " 400 K, K_(p) = 41." Find the value of "K_(p)` for each of the followingreactions at the same temperature : (i) ` 2 NH_(3) (g) hArr N_(2) (g) + 3 H_(2) (g) ` (ii) `1/2 N_(2) (g) + 3/2H_(2) (g) hArr NH_(3) (g)` (iii) `2 N_(2) (g) + 6 H_(2) (g) hArr 4 NH_(3) (g)` |
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Answer» (i) It is the reverse of the given reaction. Hence , `K_(p) = 1/41` (ii) It is obtained by dividing the given equation by 2. Hence, `K_(p) = sqrt(41)` (iii) It is obtained by multiplying the given equation by 2. Hence, `K_(p) = (41)^(2)`. |
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| 1044. |
In each of the following questions, a statement of Assertion is given followed by a corresponding statement of Reason just below it. Of the statements, mark the correct answer asAssertion. In heterogenous equilibrium , the active masses of oure solids or oure liquids are taken as constant . Reason. Pure solids and pure liquids have fixed densities and definite molecular masses.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 the true explanation of the assertion .C. If assertion 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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| 1045. |
In each of the following questions, a statement of Assertion is given followed by a corresponding statement of Reason just below it. Of the statements, mark the correct answer as Assertion . The equilibrium constant of a reaction increases if temperature is increased . Reason . The forward reaction becomes faster with increase of 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 the true explanation of the assertion .C. If assertion is true, but reason is false.D. If both assertion and reason are false . |
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Answer» Correct Answer - D Correct A and R . Efffect of temperature on equilibrium constatn K depends upon whetehr the reaction is exothermic or endother mic . |
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| 1046. |
In each of the following questions, a statement of Assertion is given followed by a corresponding statement of Reason just below it. Of the statements, mark the correct answer as Assertion. The equilibrium constant is fixed and characterstic for any given chemical reaction at a specified temperature . Reason . The composition of the final equilibrium mixture at a particular temperature depends upon the starting amount of reactants .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 the true explanation of the assertion .C. If assertion is true, but reason is false.D. If both assertion and reason are false . |
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Answer» Correct Answer - A If the starting amounts are changed , the composition of the equilibrium mixture, changes is such a way that K remains constant at constant at constant temperature . |
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| 1047. |
In each of the following questions, a statement of Assertion is given followed by a corresponding statement of Reason just below it. Of the statements, mark the correct answer as Assertion . If standard free energy change of a reaction is zero , this implies that equilibrium constant of the reaction is unity . Reason . For a reaction in equilibrium , equilibrium constant is always unity .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 the true explanation of the assertion .C. If assertion is true, but reason is false.D. If both assertion and reason are false . |
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Answer» Correct Answer - C `DeltaG^(@) = -2 *303 RT log K ` When ` DeltaG^(@) =0, log K=0 :. K=1` Hence, Assertion is correct . Correct R. For a reaction in equilibrium , the value of equilibrum constant depends upon the relative amounts of the reactants and products . |
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| 1048. |
In each of the following questions, a statement of Assertion is given followed by a corresponding statement of Reason just below it. Of the statements, mark the correct answer as Assertion. When a catalyst is added to a reaction mixture in equilibrium , the amount of the products increases . Reason . The forward reaction becomes faster on adding the catalyst .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 the true explanation of the assertion .C. If assertion is true, but reason is false.D. If both assertion and reason are false . |
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Answer» Correct Answer - D Correct A and R Both the forward and backward reaction become faster and equilibrium is not disturbed . |
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| 1049. |
The temperature at which `K_(c) and K_(p)` will have the same value for the equilibrium , ` N_(2) O_(4) (g) hArr 2 NO_(2) (g) ` isA. 0 KB. 273 KC. 1 KD. 12.18 K |
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Answer» Correct Answer - D ` K_(p) = K_(c) (RT)^(Delta n_(g)). " Here" , Delta n _(g) = 2-1 = 1. " Hence " ,` ` i.e. T= 1/R = 1/(0*0821)= 12*18 K` |
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| 1050. |
The equilibrium constant for the reactions `N_(2) + O_(2) hArr 2 NO and (ii) 2 NO + O_(2) hArr 2 NO_(2) "are "K_(1) and K_(2)` respectively, then what will be the equilibrium constant for the reaction ` N_(2) + 2 O_(2) hArr2 NO_(2) ` |
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Answer» Adding eqns. (i) and (ii), we get the required equation. Thus, (i) and (ii) are the two steps of the required equation. Hence, for the required reaction, `K= K_(1) xx K_(2) .` |
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