JEE MAIN - Chemistry (2015 (Offline) - No. 6)
The standard Gibbs energy change at 300 K for the reaction 2A $$\leftrightharpoons$$ B + C is 2494.2 J. At a given time,
the composition of the reaction mixture is [A] = 1/2, [B] = 2 and [C] = 1/2. The reaction proceeds in the: [R = 8.314 J/K/mol, e = 2.718]
reverse direction because Q > Kc
forward direction because Q < Kc
reverse direction because Q < Kc
forward direction because Q > Kc
Explanation
$$\Delta {G^ \circ } = 2494.2J$$
$$2A\,\rightleftharpoons\,B + C$$
$$R = 8.314\,J/K/mol.$$
$$e = 2.718$$
$$\left[ A \right] = {1 \over 2},\left[ B \right] = 2,$$
$$\left[ C \right] = {1 \over 2};Q = {{\left[ B \right]\left[ C \right]} \over {{{\left[ A \right]}^2}}}$$
$$ = {{2 \times 1/2} \over {{{\left( {{1 \over 2}} \right)}^2}}} = 4$$
$$\Delta {G^ \circ } = - 2.303\,\,RT\,\log \,{K_c}.$$
$$2494.2J = - 2.303 \times \left( {8.314J/K/mol} \right)$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \times \left( {300K} \right)\log {K_c}$$
$$ \Rightarrow \log \,{K_c}$$
$$ = - {{2494.2\,J} \over {2.303 \times 8.314\,J/K/mol \times 300\,K}}$$
$$ \Rightarrow \log \,{K_c} = - 0.4341;\,\,{K_c} = 0.37;\,\,Q > {K_c}.$$
$$2A\,\rightleftharpoons\,B + C$$
$$R = 8.314\,J/K/mol.$$
$$e = 2.718$$
$$\left[ A \right] = {1 \over 2},\left[ B \right] = 2,$$
$$\left[ C \right] = {1 \over 2};Q = {{\left[ B \right]\left[ C \right]} \over {{{\left[ A \right]}^2}}}$$
$$ = {{2 \times 1/2} \over {{{\left( {{1 \over 2}} \right)}^2}}} = 4$$
$$\Delta {G^ \circ } = - 2.303\,\,RT\,\log \,{K_c}.$$
$$2494.2J = - 2.303 \times \left( {8.314J/K/mol} \right)$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \times \left( {300K} \right)\log {K_c}$$
$$ \Rightarrow \log \,{K_c}$$
$$ = - {{2494.2\,J} \over {2.303 \times 8.314\,J/K/mol \times 300\,K}}$$
$$ \Rightarrow \log \,{K_c} = - 0.4341;\,\,{K_c} = 0.37;\,\,Q > {K_c}.$$
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