JEE MAIN - Chemistry (2021 - 24th February Evening Shift - No. 18)
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)
$$3HC \equiv C{H_{(g)}} \rightleftharpoons {C_6}{H_{6(l)}}$$
[Given : $${\Delta _f}{G^o}(HC \equiv CH) = - 2.04 \times {10^5}$$ J mol$$-$$1 ; $${\Delta _f}{G^o}({C_6}{H_6}) = - 1.24 \times {10^5}$$ J mol$$-$$1 ; R = 8.314 J K-1 mol$$-$$1]
$$3HC \equiv C{H_{(g)}} \rightleftharpoons {C_6}{H_{6(l)}}$$
[Given : $${\Delta _f}{G^o}(HC \equiv CH) = - 2.04 \times {10^5}$$ J mol$$-$$1 ; $${\Delta _f}{G^o}({C_6}{H_6}) = - 1.24 \times {10^5}$$ J mol$$-$$1 ; R = 8.314 J K-1 mol$$-$$1]
Answer
855
Explanation
Reaction,
$$\mathop {3HC \equiv CH(g)}\limits_{Acetylene} \to \mathop {{C_6}{H_6}(l)}\limits_{Benzene} $$
Given, $$\Delta G_f^o (CH \equiv CH) = - 2.04 \times {10^5}$$ J mol$$-$$1
$$\Delta G_f^o({C_6}{H_6}) = - 1.24 \times {10^5}$$ J mol$$-$$1
Gibb's free energy, $$\Delta G_f^o = - nRT\ln K$$
$$\Delta G_f^o = \sum {{{(\Delta G_F^o)}_P} - \sum {{{(\Delta G_F^o)}_R}} } $$
$$ - nRT\ln K = - n'RT\ln {K_p} - ( - n''RT\ln {K_f})$$
$$ \Rightarrow - RT\ln K = 1 \times ( - 1.24 \times {10^5}) - ( - 3 \times 2.04 \times {10^5}) $$
$$ \Rightarrow $$ $$- 2.303 \times R \times T\log K = 4.88 \times {10^5}$$
$$ \Rightarrow $$ $$\log K = - {{4.88 \times {{10}^5}} \over {2.303 \times 8.314 \times 273}}$$
$$ \Rightarrow $$ $$n\ln K = n'\ln {K_p} - ( - n''ln{K_f})$$
$$ \Rightarrow $$ K = 85.52
$$\Rightarrow$$ K = 855 $$\times$$ 10$$-$$1
x = 855
$$\mathop {3HC \equiv CH(g)}\limits_{Acetylene} \to \mathop {{C_6}{H_6}(l)}\limits_{Benzene} $$
Given, $$\Delta G_f^o (CH \equiv CH) = - 2.04 \times {10^5}$$ J mol$$-$$1
$$\Delta G_f^o({C_6}{H_6}) = - 1.24 \times {10^5}$$ J mol$$-$$1
Gibb's free energy, $$\Delta G_f^o = - nRT\ln K$$
$$\Delta G_f^o = \sum {{{(\Delta G_F^o)}_P} - \sum {{{(\Delta G_F^o)}_R}} } $$
$$ - nRT\ln K = - n'RT\ln {K_p} - ( - n''RT\ln {K_f})$$
$$ \Rightarrow - RT\ln K = 1 \times ( - 1.24 \times {10^5}) - ( - 3 \times 2.04 \times {10^5}) $$
$$ \Rightarrow $$ $$- 2.303 \times R \times T\log K = 4.88 \times {10^5}$$
$$ \Rightarrow $$ $$\log K = - {{4.88 \times {{10}^5}} \over {2.303 \times 8.314 \times 273}}$$
$$ \Rightarrow $$ $$n\ln K = n'\ln {K_p} - ( - n''ln{K_f})$$
$$ \Rightarrow $$ K = 85.52
$$\Rightarrow$$ K = 855 $$\times$$ 10$$-$$1
x = 855
Comments (0)
