Answer:
[tex]\delta G= -261.2kg[/tex] for the reaction under a given set of nonstandard conditions.
Explanation:
[tex]C_{2}H_6(g) + 2H_2(g)\rightleftharpoons C_2H_6(g)[/tex]
[tex]Q_p = \frac{P_c_2H_6}{P_c_2H_2\timesP_H_2}[/tex]
= [tex]\frac{3.25\times10^-2}{4.25\times4.15}[/tex]
[tex]Q_p[/tex] = [tex]4.44 \times 10^-4[/tex]
[tex]\delta G =\delta G^0 + RTlnQ_P[/tex]
= [tex]-242.1+8.314\times10^-3\times298\timesln(4.44\times10^-4)[/tex]
= [tex]\delta G= -261.2kg[/tex]
So, [tex]\delta G= -261.2kg[/tex]