Discussion Forum



1)

For a reaction , $A\rightleftharpoons P,$  the plots of [A]  and [P] with time at temperature T1  and T2 are given below.

1792019855_time.JPG

 If T2 > T1 , the correct statement (s) is /are 

     [ Assume $\triangle H^{\ominus}$ and $\triangle S^{\ominus}$ are independent of temperature and ratio of ln K at T1 to ln K at T2 is greater than T2/T1. Here  H, S, G and K are enthalpy, entropy, Gibbs energy  and equilibrium constant, respectively.)

 


A) $\triangle H^{\ominus}<0, \triangle S^{\ominus}<0$

B) $\triangle G^{\ominus}<0, \triangle H^{\ominus}>0$

C) $\triangle G^{\ominus}<0, \triangle S^{\ominus}<0$

D) $\triangle G^{\ominus}<0, \triangle S^{\ominus}>0$

Answer:

Option A,C

Explanation:

For the reaction, $A\rightleftharpoons P,$

Then  T1 <T2

     $\frac{ln K_{1}}{ln K_{2}}>\frac{T_{2}}{T_{1}}$

It shows,On increasing the temperature,  K decreases so reaction is exothermic i.e.,

 Δ H0 <  0

Besides, graph shows K >1

     So , Δ G0 <0

   Now from equation (i)

    T1 ln K1  > T2 ln K2

   -Δ G0 1  >  -Δ G02  

 Like wise

   (- Δ H0 +T1Δ S0 )> (- Δ H0 +T2Δ S0 )

or simply

                 T1Δ S 0  >T2 Δ S0

So,     (T2-T1) Δ S0 < 0

           Δ S0 <0

In other words, the increase of Δ G with an increase in temperature is possible only when Δ S0 <0.

   Hence, options (a) and (c) are correct.