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QJEEM21Chb72f: JEE Main 2021 Chemistry - Class 11 JEEM21Chb72f Solution: Assuming ideal behaviour, the magnitude of log K for the…

Assuming ideal behaviour, the magnitude of log K for the following reaction at 25 ∘^\circ C is x ×\times 10 −- 1. The value of x is ______________. (Integer answer)

3HC≡CH(g)⇌C6H6(l)3HC \equiv C{H_{(g)}} \rightleftharpoons {C_6}{H_{6(l)}}

[Given : ΔfGo(HC≡CH)=−2.04×105{\Delta _f}{G^o}(HC \equiv CH) = - 2.04 \times {10^5} J mol −- 1 ; ΔfGo(C6H6)=−1.24×105{\Delta _f}{G^o}({C_6}{H_6}) = - 1.24 \times {10^5} J mol −- 1 ; R = 8.314 J K-1 mol −- 1]

Step-by-Step Verified Solution

Reaction,

3HC≡CH(g)Acetylene→C6H6(l)Benzene\mathop {3HC \equiv CH(g)}\limits_{Acetylene} \to \mathop {{C_6}{H_6}(l)}\limits_{Benzene}

Given, ΔGfo(CH≡CH)=−2.04×105\Delta G_f^o (CH \equiv CH) = - 2.04 \times {10^5} J mol −- 1

ΔGfo(C6H6)=−1.24×105\Delta G_f^o({C_6}{H_6}) = - 1.24 \times {10^5} J mol −- 1

Gibb's free energy, ΔGfo=−nRTln⁡K\Delta G_f^o = - nRT\ln K

ΔGfo=∑(ΔGFo)P−∑(ΔGFo)R\Delta G_f^o = \sum {{{(\Delta G_F^o)}_P} - \sum {{{(\Delta G_F^o)}_R}} }

−nRTln⁡K=−n′RTln⁡Kp−(−n′′RTln⁡Kf)- nRT\ln K = - n'RT\ln {K_p} - ( - n''RT\ln {K_f})

⇒−RTln⁡K=1×(−1.24×105)−(−3×2.04×105)\Rightarrow - RT\ln K = 1 \times ( - 1.24 \times {10^5}) - ( - 3 \times 2.04 \times {10^5})

⇒\Rightarrow −2.303×R×Tlog⁡K=4.88×105- 2.303 \times R \times T\log K = 4.88 \times {10^5}

⇒\Rightarrow log⁡K=−4.88×1052.303×8.314×273\log K = - {{4.88 \times {{10}^5}} \over {2.303 \times 8.314 \times 273}}

⇒\Rightarrow nln⁡K=n′ln⁡Kp−(−n′′lnKf)n\ln K = n'\ln {K_p} - ( - n''ln{K_f})

⇒\Rightarrow K = 85.52

⇒\Rightarrow K = 855 ×\times 10 −- 1

x = 855

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