JEE Main 2021 — Chemical Equilibrium Question with Solution

For a reaction, A(g) $$\to$$ B(g)

Given, K_p (equilibrium constant) = 100

Temperature = 300 K

Pressure = 1 atm

Formula used, $$\Delta$$G$${}^{\circ }$$ = $$-$$ RT ln K_p .... (i)

Here, $$\Delta$$G$${}^{\circ }$$ = standard Gibb's free energy

R = gas constant = 8.31 J mol^$$-$$1 K^$$-$$1

Put value in Eq. (i), we get

$$\Delta$$G$${}^{\circ }$$ = $$-$$ R (300) ln 100

$$\Delta$$G$${}^{\circ }$$ = $$-$$ R (300) (2) ln (10)

$$\because$$ ln (10) = 2.3

$$\Delta$$G$${}^{\circ }$$ = $$-$$ R(300) (2) (2.3)

$$\Delta$$G$${}^{\circ }$$ = $$-$$ 1380 R

Hence, $$\Delta$$G$${}^{\circ }$$ = $$-$$ xR

$$\therefore$$ x = 1380