QJEEM21Chaa61: JEE Main 2021 Chemistry - Class 11 JEEM21Chaa61 Solution: For the reaction A (g) B (g) the value of the equilibrium…

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