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InterviewSolution
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For the reaction `N_(2)O_(4)(g)hArr2NO_(2)(g)`, the value of `K_(p)` is `1.7xx10^(3)` at `500K` and `1.7xx10^(4)` at `600K`. Which of the following is/are correct ?A. The proportions of `NO_(2)` in the equilibrium mixture is increased by decrease in pressure.B. The standard enthalpy change for the forward reaction is negativeC. Units of `K_(p)` are `atm^(-1)`D. At `500 K` the degree of dissociation of `N_(2)O_(4)` decreases by `50%` by increasing the pressure by `100%` |
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Answer» Correct Answer - A `Deltan=2-1=1` a. That is, with the decrease of pressure, reaction shifts towards right i.e. proportions of `NO_(2)` increases. Statement (a) is correct. b. Value of K increases with increase of temperature and hence reaction is endothermic, i.e., `DeltaH=+ve` Hence statement (b) is incorrect. c. `K_(p)=([p_(NO_(2))]^(2))/([p_(N_(2)O_(4))])=("atm"^(2))/("atm")="atm"` Hence statement (c) is incorrect. d. `{:(,N_(2)O_(4)(g),hArr,2NO_(2)(g)),("Initial",1 "atm",,0),("At eq.",1-alpha,,2alpha):}` `K_(p)=((2alpha)^(2))/(1-alpha)=1.7xx10^(3)` at `500 K` `[(1-alpha~~1), "since" alpha "is small"]` `4alpha^(2)=1.7xx10^(3)` `alpha_(1)=sqrt((1.7xx10^(3))/4)` `alpha=0.206xx10^(2)` When the pressure was `1` atm, now let it be `2` atm `(100% "increase")`. `{:(,N_(2)O_(4)(g),hArr,2NO_(2)(g)),("Initial",2,,0),("At eq.",2-alpha,,2alpha):}` `K_(p)=((2alpha)^(2))/(2-alpha)=1.7xx10^(3)` at `500 K` `[2-alpha~~2, "since" alpha "is small"]` `(4alpha^(2))/2=1.7xx10^(3)` `alpha_(2)=sqrt((1.8xx10^(3))/2)` `alpha_(2)/alpha_(1)=sqrt((1.7xx10^(3)xx4)/(2xx1.7xx10^(3)))=sqrt(2)=1.4` `:.` When the pressure is increased `100%` the decrease in `alpha` is `1.4` times which is not `50%`. Hence statement is wrong. e. `K_(p)` at `600 K=1.78xx10^(4)` `N_(2)O_(4)(g) hArr 2NO_(2)(g)` `Deltan=2-1=1` Since, by decrease of pressure reaction goes forward, i.e., more of `N_(2)O_(4)` will dissociate. It means by decreasing pressure dissociation of `N_(2)O_(4)` increases. Hence, the statement is wrong. |
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