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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.
| 901. |
For the reaction `CaCO_3(s)hArr CaO(s)+CO_2(g)` , the pressure of `CO_2` (g) depends on :A. the mass of `CaCO_3(s)`B. the mass of CaO(s)C. the masses of both `CaCO_3(s)` and CaO(s)D. temperature of the system |
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Answer» Correct Answer - D |
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| 902. |
What percent of `CO_2` in air is just sufficient to prevent loss in weight when `CaCO_3` is heated at `100^@C`? (Equilibrium contant K for `CaCO_3(s) hArr CaO(s)+CO_2(g)` is 0.0095 atm at `100^@C`)A. `0.95%`B. `0.29%`C. `0.05%`D. `0.71%` |
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Answer» Correct Answer - A |
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| 903. |
For the reaction ` : CaCO_(3)(s) hArr CaO(s)+CO_(2)(g),K_(p)=1.16atm` at `800^(@)C` . If `20g` of `CaCO_(3)` were kept in a 10 litre vessel at `800^(@)C`, the amount of `CaCO_(3)` remained at equilibrium is `:`A. `34%`B. `64%`C. `46%`D. none |
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Answer» Correct Answer - 1 `{:(CaCO_(3)(s),hArr,CaO(s),+,CO_(2)(g)),(0.2-x,,x,,x):}` `K_(p)=P_(co2)=1.16` `x=` moles of `CO_(2)=(KV)/(RT)` Remaining mass of `CaCO_(3)=(0.2-x)=100g`. |
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| 904. |
Ammonia gas at 15 atm is introduced in a rigid vessel at 300 K. At equilibrium the total pressure of the vessel is found to be 40.11 atm at `300^@C`.The degree of dissociation of `NH_3` will be :A. 0.6B. 0.4C. unpredictableD. None of these |
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Answer» Correct Answer - B |
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| 905. |
For the reaction : `CaCO_3(s) hArr CaO(s)+CO_2(g), K_p=1.6` atm at `800^@C`.if 20 g of `CaCO_3` were kept in a 10 litre vessel at `800^@C`, the amount of `CaCO_3` that remained at equilibrium is :A. `34%`B. `64%`C. `46%`D. None of these |
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Answer» Correct Answer - D |
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| 906. |
statement-1 : Ammonia at a pressure of 10 atm and `CO_(2)` at a pressure of 20atm are introduced into an evacuated chamber. If `K_(P)` for the reaction. `NH_(2)COONH_(4)hArr2NH_(3)(g)+CO(g)` is `2020atm^(3)` the total pressure after a long time is less than 30atm. statement-2 : Equilibrium can be attained from both directions.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statment-2is True ,Statement-2is not a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - D |
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| 907. |
If dissociation for reaction, `PC1_(5)hArrPC1_(3) + C1_(2)` is `20%` at 1 atm pressure. Calculate `K_(c)`.A. `0.04`B. `0.05`C. `0.07`D. `0.06` |
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Answer» Correct Answer - B `K_(c)=([PC1_(3)][C1_(2)])/([PC1_(5)])=([(20)/(100)]xx[(20)/(100)])/([(80)/(100)])` `= (0.2xx0.2)/(0.8)=(0.04)/(0.8)=0.05` |
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| 908. |
The degree of dissociation of `PC1_(5)` `(alpha)` obeying the equilibrium, `PC1_(5)hArrPC1_(3) + C1_(2)`, is approximately related to the pressure at equilibrium by (given `alpha ltlt 1`) :A. `alpha prop P`B. `alpha prop 1/sqrtP`C. `a prop 1/P^2`D. `alpha prop 1/P^4` |
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Answer» Correct Answer - B |
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| 909. |
Ammonia gas at 15 atm is introduced in a rigid vessel at 300 K. At equilibrium the total pressure of the vessel is found to be 40.11 atm at `300^@C`.The degree of dissociation of `NH_3` will be :A. `0.6`B. `0.4`C. unpredictableD. None of these |
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Answer» Correct Answer - B `2NH_(3)hArrN_(2) + 3H_(2)` `T = 300K` 15 atm `T = 573 K` `(15)/(300) (573)` atm `(15)/(300) (573)` `[1 + alpha) = 40.11` `implies` `alpha = 0.4` |
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| 910. |
`K_(c) = 9` for the reaction, `A + B hArrC + D`. If A and B are taken in equal amounts, then amount of C in equilibrium is:A. 1B. `0.25`C. `0.75`D. None of these |
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Answer» Correct Answer - C `{:(,A,+,B,hArr C,+,D),("Initial",1,1,,0,,0),("At equilibrium",(1-x),,(1-x),x,,x):}` `:.` `K_(c) = ([C][D])/([A][B]) =9` `:.` `(x.x)/((1-x)^(2)) =9` or `x^(2)=9+9x^(2)-18 x` or `8x^(2)-18x+9=0` `:.` `x=(3)/(2)` or `(3)/(4)` Hence, among the given options, option (c), i.e., `0.75` is correct. |
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| 911. |
Assertion: The dissociation of `PC1_(5)` decreases on increasing pressure. Reason: An increase in pressure favours the forward reaction.A. If both assertion and reason are true and the reason is the true explanation of the assertion.B. If both the assertion and reason are true but the reason is not the correst explanation of assertion.C. If the assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - C `{:(PCl_(5),hArr,PCl_(3),+,Cl_(2)),(1,,0,,0),(1-alpha,,alpha,,alpha):}` `K_(p) = [(alpha^(2))/(1 - alpha)][(P)/(1 + alpha)]^(1)` An increase in pressure will decrease `alpha` , to have `K_(p)` constant and thus backward reaction occurs. |
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| 912. |
Assertion: The dissociation of `PC1_(5)` decreases on increase. Reason: An increase in pressure favours the forward reaction.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - C `{:(PCl_(5),hArr,PCl_(3),+,Cl_(2)),(1,,0,,0),(1-alpha,,alpha,,alpha):}` `K_(p) = [(alpha^(2))/(1-alpha)] [(P)/(1+alpha)]^(1)` An increase in pressure will decrease `alpha`, to have `K_(p)` constant and thus backward reaction occurs. |
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| 913. |
The rate of forward reaction is two times that of reverse reaction at a given temperature and identical concentration. `K_(equilibrium)` isA. `2.5`B. `2.0`C. `0.5`D. `1.5` |
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Answer» Correct Answer - B The rate of forward reaction is two times that of reverse reaction at a gives temperature and identical concetration `K_(equilibrium)` is 2 because the reaction is revessible. So `K = (K_(1))/(K_(2)) = (2)/(1) = 2` |
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| 914. |
Assertion: The melting point of solid (except ice) increases with increase in pressure. Reason: An increasion in pressure favours the change where volume decreases.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - A Solide `hArr` Liquid `V_(L) gt V_(S)` , increase in pressure leads to melting of ice as it favours the change showing decrease in volume. |
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| 915. |
Why does ice melt showly at higher altitudes? |
| Answer» Melting of ice in endothermic process accomanied by decreases of volume. At altitudes, pressure as well as temperature is low. In the equibrium, Ice `harr` Water, both the factors favour backward process. Hence the melting is slow. | |
| 916. |
When 1 mole of carbon is converted into 1 mole of `CO_(2)`, the heat liberated is same :A. irrespective of whether the volume is kept constant or pressure is kept constantB. irrespective of the temperature at which there reaction is carried outC. whether the carbon is in the form of diamond or graphiteD. whether the carbon is in gaseous state or solid state. |
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Answer» Correct Answer - A |
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| 917. |
The increase of pressure on ice water system at constant temperature will lead to :A. a shift of the equilibrium in the forward directionB. a decrease in the entropy of the systemC. an increase in the Gibbs energy of the systed no effect on the equilibriumD. no effect on the equilibrium |
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Answer» Correct Answer - A |
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| 918. |
`PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)` In given equilibrium reaction, volume of container is increased by mixing inert gas at constant temperature :A. equilibrium will shift in forward direction and concentration of `Cl_(2)(g)` will decreaseB. equilibrium will shift in forward direction and concentration of `Cl_(2)(g)` will increaseC. equilibrium will shift in backward direction and concentration of `Cl_(2)(g)` will decreaseD. equilibrium will whift in backward direction and concentration of `Cl_(2)(g)` will increase |
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Answer» Correct Answer - A |
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| 919. |
The following two reactions: i. `PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)` (ii) `COCl_(2)(g) hArr CO(g)+Cl_(2)(g)` are simultaneously in equilibrium in a container at constant volume. A few moles of `CO(g)` are later introduced into the vessel. After some time, the new equilibrium concentration ofA. `PCl_5` will remain unchangedB. `Cl_2` will be greaterC. `PCl_5` will become lessD. `PCl_5` will become greater |
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Answer» Correct Answer - C |
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| 920. |
For the reaction, `PCl_(3)(g)+Cl_(2)(g) hArr PCl_(5)(g)`, the position of equilibrium can be shifted to the right by:A. increasing the temperatureB. Compressing the gaseous mixtureC. Increasing the volume of the gaseous mixtureD. Adding `Cl_(2(g))` to the equilibrium mixture at a constant volume |
| Answer» Correct Answer - (a,c) | |
| 921. |
When `NaNO_(3)` is heated in a closed vessel, oxygen is liberated and `NaNO_(2)` is left behind. At equilibrium,A. addition of `NaNO_(2)` favours reverse reactionB. addition of `NaNO_(3)` favours forward reactionC. increasing temperature favours forward reactionD. increasing pressure favours reverse reaction |
| Answer» Correct Answer - (c,d) | |
| 922. |
When `NaNO_(3)` is heated in a closed vessel, oxygen is liberated and `NaNO_(2)` is left behind. At equilibrium, which are correctA. Addition of `NaNO_(2)` favours reverse reactions.B. Addition of `NaNO_(2)` favours forward reactions.C. Increasing temperature favours forward reactionD. Increasing pressure reverse reaction. |
| Answer» Correct Answer - C::D | |
| 923. |
The density of an equilibrium mixture of `N_(2)O_(4)` and `NO_(2)` at 1 atm and 373.5K is 2.0 g/L. Calculate `K_(C)` for the reaction `N_(2)O_(2)(g) iff 2NO_(2)(g)` |
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Answer» Correct Answer - 2 |
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| 924. |
For the reaction `N_(2(g)) + O_(2(g))hArr2NO_((g))`, the value of `K_(c)` at `800^(@)C` is `0.1`. When the equilibrium concentrations of both the reactants is `0.5` mol, what is the value of `K_(p)` at the same temperatureA. `0.5`B. `0.1`C. `0.01`D. `0.025` |
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Answer» Correct Answer - B `N_(2(g)) + O_(2(g))hArr2NO_((g))` `K_(c) = 0.1, K_(p) = K_(c)(RT)^(Delta n)` `Delta n = 0, K_(p) = K_(c) = 0.1` |
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| 925. |
The density of an equilibrium mixture of `N_(2)O_(4)` and `NO_(2)` at `1` atm is `3.62 g L^(-1)` at `288 K` and `1.84 g L^(-1)` at `348 K`. Calculate the entropy change during the reaction at `348 K`. |
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Answer» `N_(2)O_(4) hArr 2NO_(2)` For `K_(p)`, proceed as follows: `PV=nRT=w/m_(mix) RT` `rArr m_(mix)=w/Vxx(RT)/(P)=(dRT)/(P)` `=3.62xx0.082xx288=85.6` Let a mol of `N_(2)O_(4)` and `(1-a)` mol of `NO_(2)` exist at equilibrium. `:. axx92+(1-a)xx46=85.6` `:. a=0.86` `:. n_(N_(2)O_(4))=0.86, n_(NO_(2))=0.14 "mol"` `K_(p)=(0.14xx0.14)/(0.86)xx[1/1]^(1)=0.0228` atm at `288 K` Case II `m_(mix)=(dRT)/(P)=1.84xx0.0821xx348=52.57` Let a mol of `N_(2)O_(4)` and `(1-a)` mol of `NO_(2)` exist at equilibrium. `:. axx92+(1-a)xx46=52.57` `:. a=0.14` `:. n_(N_(2)O_(4))=0.14, n_(NO_(2))=0.86` `:. K_(p)=(0.86xx0.86)/(0.14)[1/1]^(1)=5.283` "atm" at `348 K` `log_(10)=(K_(P_(2))/(K_(P_(1))))=(DeltaH)/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]` `rArr "log"_(10) 5.283/0.0228=(DeltaH)/(2xx2.303)[(348-288)/(348xx288)]` `:. DeltaH=181956 cal=18.196` kcal `DeltaG=-2.303 RT log K_(p)` `=-2.303xx2xx348xxlog 5.283` `=-1158.7 cal` `DeltaS=(DeltaH-DeltaG)/(T)=(18195.6+1158.7)/(348)=55.62 cal` |
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| 926. |
Density of equilibrium mixture of `N_(2)O_(4)` and `NO_(2)` at `1 atm` and `384 K` is `1.84 g dm^(-3)`. Calculate the equilibrium constant of the reaction. `N_(2)O_(4)hArr2NO_(2)` |
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Answer» We know `Pm=dRT` `1xxm=1.84xx0.0821xx384` `m=29xx2` Vapour density (d) at equilibrium `=29` Initial vapour density`=M//2=92//2=46` therefore, degree of dissociation is: `x=(D-d)/((n-1)d)=(46-29)/(29)=0.586` For reaction `N_(2)O_(4)hArr2NO_(2)` `{:(t=0,1,,0,),(t_(eq),1-x,,2x,"Total moles"=1+x):}` `p_(N_(2)O_(4))=(1-x)/(1+x)xxP, p_(NO_(2))=(2x)/(1+x)xxP` `K_(p)=(4x^(2)P)/(1-x^(2))=(4xx(0.586)^(2)xx1)/(1-(0.586)^(2))=2.09 "atm"` |
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| 927. |
For the reaction `NH_(3)(g)hArr(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g)` Show that the degree of dissociation of `NH_(3)` is given as `alpha=[1+(3sqrt(3))/4p/K_(p)]^(-1//2)` where p is equilibrium pressure. If `K_(p)` of the above reaction is `78.1 atm` at `400^(@)C`, calculate `K_(c )`. |
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Answer» `NH_(3)(g)hArr(1)/(2)N_(2)+(3)/(2)H_(2)(g)` Total moles `{:(t=0,1,0,0,1),(t_(eq),1-alpha,alpha//2,3alpha//2,1+alpha):}` `p_(i) {(1-alpha)/(1+alpha)}p, {(alpha)/(2(1+alpha))}p, {(3alpha)/(2(1+alpha))}p` `K_(p)=((P_(N_(2)))^(1//2)(P_(H_(2)))^(3//2))/((P_(NH_(3))))` `=([alpha/(2(1+lpha))p]^(1//2)[(3alpha)/(2(1+alpha))p]^(3//2))/([(1-alpha)/(1+alpha)p])=(palpha^(2)sqrt(27))/(4(1-alpha^(2)))` Solving for `alpha`, we get `alpha=[1+(3sqrt(3))/4p/K_(p)]^(1//2)` `K_(c )` can be caiculate by using `K_(p)=K_(c )(RT)^(Deltan)` `K_(p)=78.1, T=673, Deltan=1 K_(c )=K_(p)(RT)^(-1)=78.1/((0.082xx673))` `K_(c )=78.1/55.18=1.415` |
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| 928. |
Using the Gibbs energy change, `Delta G^(@)=+ 63.3 kJ`, for the following reaction, `Ag_(2)CO_(3)hArr2Ag^(+)(aq)+CO_(3)^(2-)` the `K_(sp)` of `Ag_(2)CO_(3)(s)` in water at `25^(@)C` is `(R=8.314 JK^(-1)mol^(-1))`A. `3.2xx10^(-26)`B. `8.0xx10^(-12)`C. `2.9xx10^(-3)`D. `7.9xx10^(-2)` |
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Answer» Correct Answer - B `DeltaG^@` is related to `K_sp` by the equation. `DeltaG^@=-2.303"RT log "K_sp` Given, `Deltag^@=+63.3kJ` `=63.3xx10^3J` Thus, substance `DletaG^@=63.3xx10^3J`. `R=8.314JK^(-1)mol^(-1)` and `T=298K[25+273K]` from the above equation we get, `63.3xx10^3=--2.303xx8.314xx298logK_sp` `therefore" "log K_sp=-11.09` `rArrK_sp ="anti log "(-11.09)` `K_sp=8.0xx10^(-12)` |
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| 929. |
Consider the reaction, `2CI_(2)(g)+2H_(2)O(g)hArr4HCI(g)+P_(2)(g) DeltaH^(@)=+113KJ` The four gases, `CI_(2),H_(2)O,HCI "and" O_(2)`, are mixed and the reaction is allowed to come to equilibrium. Each operation is to be considered separately. Temperature and volume are constant unless stted otherwise. Report the number of operations in the left column which lead to increase in the equilibrium value of the quantity in the right column. (a) Increasing the volume of the container Number of moles of `H_(2)O` (b) Adding `O_(2)` Number of moles of `H_(2)O` (c) Adding `O_(2)` Number of moles of `HCI` (d) Decreasing the volume of the container Number of moles of `CI_(2)` (e) Decreasing the vollume of the container Partial pressure of `CI_(2)` (f) Decreasing the volume of the container K_(C)` (g) Raising the temperature Concentration of HCI` (i) Adding `He` Number of moles of `HCI` (j) Adding catalyst Number of moles of `HCI` |
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Answer» Correct Answer - 5 (a) decrease (b) increase (c) decrease (d) increase (e) increase (f) no change (g) increase (h) increase (i) no change (j) no change |
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| 930. |
In the dissociation of HI, `20%` of HI is dissociated at equilibrium. Calculate `K_(p)` for `HI(g) hArr 1//2 H_(2)(g)+1//2 I_(2)(g)` |
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Answer» Correct Answer - A::B `{:(,2HI,hArr,1/2 H_(2),+,1/2 I_(2)),("Initial",1,,0,,0),("moles at Eq",(1-alpha),,alpha//2,,alpha//2):}` where `alpha` is degree of dissociation and volume of container is VL. `K_(p)=K_(c)((alpha/(2V))^(1//2)(alpha/(2V))^(1//2))/(((1-alpha))/V)` `K_(p)=K_(c)=alpha/(2(1-alpha))` `alpha=0.2` `K_(p)=K_(c)=0.2/(2(1-0.2))` `K_(p)=K_(c)=0.125` |
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| 931. |
The degree of dissociation of `I_(2)` "mole"cule at `1000^(@)C` and under `1.0 atm` is `40%` by volume. If the dissociation is reduced to `20%` at the same temperature, the total equilibrium pressure on the gas will be:A. `1.57 atm`B. `2.57 atm`C. `3.57 atm`D. `4.57 atm` |
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Answer» Correct Answer - d |
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| 932. |
If two gases `AB_(2)` and `B_(2)C` are mixed, following equilibria are readily established: `AB_(2)(g)+B_(2)C(g)rarr AB_(3)(g)+BC(g)`, `BC(g)+B_(2)C(g) rarr B_(3)C_(2)(g)` If the reaction is started only with `AB_(2)` with `B_(2)C`, then which of the following us necessarily true at equilibrium?A. `[AB_(3)]_(eq)=[BC]_(eq)`B. `[AB_(2)]_(eq)=[B_(2)C]_(eq)`C. `[AB_(3)]_(eq)gt [B_(3)C_(2)]_(eq)`D. `[AB_(3)]_(eq)gt [BC]_(eq)` |
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Answer» Correct Answer - C::D `AB_(2)(g)+B_(2)(g)hArrAB_(3)(g)+BC(g)` `{:(x,y,,0,0),(x-z,y-z,,z,z):}` `[AB_(3)]=z,[B_(3)C_(2)]=a(z gt a)` `[BC]=z-a lt z` `BC(g)+B_(2)ChArrB_(3)C_(2)(g)` `{:(z,,y-z,,),(z-a,,y-z-z,,a):}` |
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| 933. |
For the synthesis of ammonia, `K_c` is 1.2 at `375^@C,N_2(g)+3H_2(g)hArr 2NH_3(g)` What is `K_p` at this temperature ?A. `4.1xx10^(-8)`B. `4.2xx10^(-4)`C. `1.3xx10^(-3)`D. `3.4xx10^(3)` |
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Answer» Correct Answer - B |
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| 934. |
For which reaction will `K_p` be larger than `K_c` at `25^@C` ?A. `CO_2(g)+C(s)to2CO(g)`B. `2NO_2(g)toN_2O_4(g)`C. `H_2(g)+F_2(g)to2HF(g)`D. `O_3(g)+NO(g)toNO_2(g)+O_2(g)` |
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Answer» Correct Answer - A |
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| 935. |
log `K_p/K_c`+log RT=0 is a relationship for the reaction :A. (A) `PCI_(5)hArrPCI_(3)+CI_(2)`B. (B) `2SO_(2)+O_(2)hArr2SO_(3)`C. (C) `H_(2)+1_(2)hArr2HI`D. (D) `N_(2)+3H_(2)hArr2NH_(3)` |
| Answer» Correct Answer - B | |
| 936. |
For the equilibrium system `2HX(g) hArr H_(2)(g)+X_(2)(g)` the equilibrium constant is `1.0xx10^(-5)`. What is the concentration of HX if the equilibrium concentration of `H_(2)` and `X_(2)` are `1.2xx10^(-3)` M, and `1.2xx10^(-4)` M respectively?A. `12xx10^(-4) M`B. `12xx10^(-3) M`C. `12xx10^(-2) M`D. `12xx10^(-1) M` |
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Answer» Correct Answer - C `{:(,2HX,hArr,H_(2)(g),+,X_(2)(g)),("At equilibrium",?,,1.2xx10^(-3) M,,1.2xx10^(-4) M):}` `K=([H_(2)][X_(2)])/([HX]^(2))` `10^(-5)=(1.2xx10^(-3)xx1.2xx10^(-4))/([HX]^(2))` `[HX]=sqrt((1.2xx1.2xx10^(-7))/(10^(-5)))` `=1.2xx10^(-1)` `=12xx10^(-2) M` |
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| 937. |
A reversible reaction is one whichA. (A) Proceeds in one directionB. (B) Proceeds in both directionsC. (C) Proceeds spontaneouslyD. (D) All the statements are wrong |
| Answer» Correct Answer - B | |
| 938. |
If `DeltaG^(@)` for the reaction given below is `1.7 kJ`, the equilibrium constant of the reaction, `2HI_((g))hArrH_(2(g))+I_(2(g))` at `25^(@)C` is :A. `24.0`B. `3.9`C. `2.0`D. `0.5` |
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Answer» Correct Answer - D `Delta G^(@) = - 2.303xx8.314xx10^(-3)xx298 log K_(p)` `1.7 = - 2.303xx8.314xx10^(-3)xx298xxlog K_(p)` `K_(p) = 0.5` |
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| 939. |
The equilibrium `N_(2) (g) + O_(2) (g)hArr 2NO (g)` is established in a reaction vessel of `2.5` L capacity. The amounts of `N_(2)` and `O_(2)` taken at the start were respectively 2 moles and 4 moles. Half a mole of nitrogen has been used up at equilibrium. The molar concentration of nitric oxide is:A. 0.2B. 0.4C. 0.6D. 0.1 |
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Answer» Correct Answer - B |
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| 940. |
The equilibrium `N_(2) (g) + O_(2) (g)hArr 2NO (g)` is established in a reaction vessel of `2.5` L capacity. The amounts of `N_(2)` and `O_(2)` taken at the start were respectively 2 moles and 4 moles. Half a mole of nitrogen has been used up at equilibrium. The molar concentration of nitric oxide is:A. `0.2`B. `0.4`C. `0.6`D. `0.1` |
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Answer» Correct Answer - B `{:(,N_(2) +, O_(2),hArr,2NO),("Initial",2"moles",4"moles",,),("At.Eq.",2-(1)/(2),4-(1)/(2),,2xx(1)/(2)=1"mol"):}` Molar concentration of NO at equilibrium `=(1)/(2.5) =0.4` |
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| 941. |
The effect of adding krypton `(Kr)` gas on position of equilibrium, keeping the volume of the system constant isA. If `Deltan=0`, backward reaction is favoured.B. If,`Deltan=+ve,` forward reaction is favouredC. If `Deltan=-ve,` forward reaction is favouredD. No effect watever be the value of `Deltan` |
| Answer» Correct Answer - D | |
| 942. |
Calculate `DeltaG^(Theta)` for the conversion of oxygen to ozone, `((3)/(2)) O_(2)(g) hArr O_(3)(g) at 298 K`, of `K_(p)` for this conversion is `2.47 xx 10^(-29)`.A. 163 kJ `mol^(-1)`B. `2.4xx10^(2) kJ mol^(-1)`C. `1.63` kJ `mol^(-1)`D. `2.38xx10^(6)kJ mol^(-1)` |
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Answer» Correct Answer - A As we know that, `Delta G^(@) = - 2.303 RT log K_(p)` Therefore, `Delta G^(@) = - 2.303xx(8.314)xx(298)` `(log 2.47xx10^(-29))` `Delta G^(@) = 16,3000 J mol^(-1) = 163 kJ mol^(-1)` |
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| 943. |
Calculate `DeltaG^(Theta)` for the conversion of oxygen to ozone, `((3)/(2)) O_(2)(g) hArr O_(3)(g) at 298 K`, of `K_(p)` for this conversion is `2.47 xx 10^(-29)`.A. `163 Kjmol^(-1)`B. `2.4xx10^(2)Kjmol^(-1)`C. `1.63KJmol^(-1)`D. `2.38xx10^(6)Kjmol^(-1)` |
| Answer» Correct Answer - A | |
| 944. |
The effect of temperature on equilibrium constant is expressed as`(T_(2)gtT_(1))` log`K_(2)//K_(1)=(-DeltaH)/(2.303)[(1)/(T_(2))-(1)/(T_(1))]`. For endothermic, false statement isA. `[(1)/(T_(2))-(1)/(T_(1))]="positive"`B. `DeltaH="positive"`C. log`K_(2)gt"log"K_(1)`D. `K_(2)gtK_(1)` |
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Answer» Correct Answer - A `"log K_(2)//"logK_(1)=(-DeltaH)/(2.303)[(1)/(T_(2))-(1)/(T_(1))] rArr DeltaH="positive"` |
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| 945. |
At room temperature, the equilibrium constant for the reaction P + Q `hArr` R + S was calculated to be `4.32` . At `425^(@)C` the equilibrium constant became `1.24xx10^(-2)` . This indicates that the reactionA. is exothermicB. is endothermicC. is difficult to predictD. no relation between `DeltaH "and" K` |
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Answer» Correct Answer - A At room temperature, `K=4.32` and at `425^(@)C`, equilibrium constant become `1.24xx10^(-4)` i.e. it is decreases with increase in temperature. So, it is exothermic reaction. |
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| 946. |
At room temperature, the equilibrium constant for the reaction P + Q `hArr` R + S was calculated to be `4.32` . At `425^(@)C` the equilibrium constant became `1.24xx10^(-2)` . This indicates that the reactionA. is exothermicB. is endothermicC. is difficult to predictD. no relation between `Delta H` and K |
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Answer» Correct Answer - A At room temperature, `K = 4.32` and at `425^(@)C`, equilibrium constant becomes `1.24xx10^(-2)` i.e., it is decreases `alpha` with increase in temerature. So, it is exothermic reaction. |
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| 947. |
The efffect of temperature on equilibrium consatant is expressed as, `log[(K_(2))/(K_(1))]=(-DeltaH)/(2.303)[(1)/(T_(2))-(1)/(T_(1))],(T_(2)gt T_(1))` For endothermic reaction false statement is : (d) `K_(2)K_(1)`.A. `(T_(2)gt T_(1))` =positiveB. `DeltaH`=positiveC. log`K_(2) gt log K_(1)`D. `K_(2)K_(1)`. |
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Answer» Correct Answer - A |
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| 948. |
At room temperature, the equilibrium constant for the reaction P + Q `hArr` R + S was calculated to be `4.32` . At `425^(@)C` the equilibrium constant became `1.24xx10^(-2)` . This indicates that the reactionA. is exothermicB. is endothermicC. is difficult to predictD. no reaction between `Delta and K` |
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Answer» Correct Answer - A |
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| 949. |
The decomposition of solid ammonium carbamate, `(NH_(4))(NH_(2)CO_(2))`, to gaseous ammonia and carbon dioxide is an endothermic reaction. `9NH_(4))(NH_(2)CO_(2)))(s)hArr2NH_(3)(g)+CO_(2)(g)` (a) When solid `(NH_(4))(NH_(2)CO_(2))` is introduced into and evacuated flask at `25^(@)C`, the total pressure of gas at equilibrium is `0.3` atm. What is the value of `K_(p) "at" 25^(@)C`? (b) Given that the decomposition reaction is at equilibrium. how would the following changes affect the total quatity of `NH_(3)` in the flask once equilibrium is re-established? (i) Adding `CO_(2)` (ii) Adding `(NH_(4))(NH_(2)CO_(2))` (iii) Removing `CO_(2)` (iv) Increasing the total volume (v) Adding neon (vi) Increasing the temperature |
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Answer» Correct Answer - (a) `2.31xx10^(-4)` (b) (i) decrease (ii) no change (iii) increase (iv) increase (v) no change (vi) increase `(NH_(4))(NH_(2)CO_(2))(S)hArr2NH_(3)(g)+CO_(2)(g)` 2P P Total pressure`=2P+P=0.3` `P=0.1`atm `K_(P)=(2P)^(2)P=4P^(3)` `=4xx10^(-3) atm^(3)` |
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| 950. |
If low pressure and low temperature are the favourable conditons for the reaction: `aA+bBhArr cC+dD` then the true statements will be :A. (a+b) lt (c+d) and `DeltaH=+X`B. `(a+b) gt (c+d) and DeltaH=+X`C. `(a+b) lt (c+d) and DeltaH=-X`D. no reaction between `Delta and K_(eq)` |
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Answer» Correct Answer - C |
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