Discussion Forum

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

 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.)

Answer:

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 H^{0 <  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.