Assuming ideal behaviour, the magnitude of $\log \mathrm{K}$ for the following reaction at $25^{\circ} \mathrm{C}$ is $x \times 10^{-1}$. The value of $x$ is $\_\_\_\_$ (integer answer) $$3 \mathrm{HC} \equiv \mathrm{CH}_{(\mathrm{g})} \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{6(\mathrm{l})}$$ [Given: $\Delta_{\mathrm{f}} \mathrm{G}^{\circ}(\mathrm{HC} \equiv \mathrm{CH})=-2.04 \times 10^{5} \mathrm{~J} \mathrm{~mol}^{-1}$; $\Delta_{\mathrm{f}} \mathrm{G}^{\circ}\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)=-1.24 \times 10^{5} \mathrm{~J} \mathrm{~mol}^{-1}$: $\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$]

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Chemical Thermodynamics

Assuming ideal behaviour, the magnitude of

lo g K

for the following reaction at

2 5^{\circ} C

x \times 1 0^{- 1}

. The value of

x

____

(integer answer)

3 HC \equiv CH_{(g)} ⇌ C_{6} H_{6 (l)}

[Given:

Δ_{f} G^{\circ} (HC \equiv CH) = - 2.04 \times 1 0^{5} J mol^{- 1}

;

Δ_{f} G^{\circ} (C_{6} H_{6}) = - 1.24 \times 1 0^{5} J mol^{- 1}

R = 8.314 JK^{- 1} mol^{- 1}

]

Answer: 855

Benzene is the only product and acetylene the only reactant, and elements in their standard states contribute nothing, so

Δ_{r} G^{\circ} = Δ_{f} G^{\circ} (C_{6} H_{6}) - 3 Δ_{f} G^{\circ} (C_{2} H_{2}) = (- 1.24 \times 1 0^{5}) - 3 (- 2.04 \times 1 0^{5}) = 4.88 \times 1 0^{5} J mo l^{- 1}

The equilibrium constant follows from

Δ_{r} G^{\circ} = - 2.303 R T lo g K

T = 298

lo g K = - \frac{4.88 \times 1 0 ^{5}}{2.303 \times 8.314 \times 298} = - 85.5

The magnitude of

lo g K

85.5

, which written as

x \times 1 0^{- 1}

gives

x = 855

JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Integer answer