The gas phase reaction $2 \mathrm{~A}(\mathrm{~g}) \rightleftharpoons \mathrm{A}_{2}(\mathrm{~g})$ At 400 K has $\Delta \mathrm{G}^{\circ}=+25.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The equilibrium constant $\mathrm{K}_{\mathrm{C}}$ for this reaction is ______ $\times 10^{-2}$. (Round off to the nearest integer). [Use: R = $8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}, \ln 10=2.3, \log _{10} 2=0.30$, 1atm = 1 bar] [antilog $(-0.3)=0.501$]

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Chemical Thermodynamics

The gas phase reaction

2 A (g) ⇌ A_{2} (g)

At 400 K has

Δ G^{\circ} = + 25.2 kJ mol^{- 1}

.
The equilibrium constant

K_{C}

for this reaction is ______

\times 1 0^{- 2}

.
(Round off to the nearest integer).
[Use: R =

8.3 J mol^{- 1} K^{- 1}, ln 10 = 2.3, lo g_{10} 2 = 0.30

, 1atm = 1 bar]
[antilog

(- 0.3) = 0.501

]

Answer: 166

Using the relationship between Gibbs free energy and equilibrium constant:

Δ G^{\circ} = - R T ln K_{p}

25200 = - 8.3 \times 400 \times ln K_{p}

ln K_{p} = - \frac{25200}{3320} = - 7.59

Converting to log base 10:

lo g K_{p} = \frac{- 7.59}{2.3} = - 3.3

K_{p} = 1 0^{- 3.3} = 1 0^{- 3} \times 1 0^{- 0.3} = 0.001 \times 0.501 = 5.01 \times 1 0^{- 4}

For the reaction

2 A ⇌ A_{2}

Δ n = - 1

K_{p} = K_{c} (R T)^{Δ n}

K_{c} = K_{p} \times (R T) = 5.01 \times 1 0^{- 4} \times 0.083 \times 400 = 5.01 \times 1 0^{- 4} \times 33.2 = 1.66 \times 1 0^{- 2}

Rounding to nearest integer:

K_{c} = 2 \times 1 0^{- 2}

. Answer: 2

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Integer answer