Question

$\Delta G^o$ at ${25}^{\circ }C$ for the reaction $\frac{1}{2}{N}_2\left(g\right)+\frac{3}{2}{H}_2\rightleftharpoons N{H}_3\left(g\right)$ is $-16.5kJmo{l}^{-1}$. Hence, thermodynamic equilibrium constant is:

• A. $1.08$
• B. $7.80\times {10}^2$
• C. $4.57\times {10}^6$
• D. $7.98\times {10}^{34}$

Answer 1

Answer: (B)

$7.80\times {10}^2$

$\Delta U=-RTln{K}_p$

$K_p=e^{-\Delta u/RT}$

$=e^{-(-16.5KJ/mol)/8.314\times 298}$

$=e^{16.5\times {10}^3/8.314\times 298}$

$=e^{6.659}=779$

$=7.79\times {10}^2=7.8\times {10}^2$ .

Thus answer is option $B$.