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(C) The Van't Hoff equation is given by: $\log \frac{K_2}{K_1} = \frac{\Delta H}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} ight]$Given:$K_1 = 1.2

Vedclass · Accepted Answer

$-1652.8$

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(C) The Van't Hoff equation is given by: $\log \frac{K_2}{K_1} = \frac{\Delta H}{2.303 R} \left[ \frac{T_2 - T_1}{T_1 T_2} ight]$Given:$K_1 = 1.2 	imes 10^{-4}$ at $T_1 = 300 \ K$$K_2 = 0.60 	imes 10^{-4}$ at $T_2 = 400 \ K$$R = 1.98 \ cal \ K^{-1} \ mol^{-1}$Substituting the values:$\log \left( \frac{0.60 	imes 10^{-4}}{1.2 	imes 10^{-4}} ight) = \frac{\Delta H}{2.303 	imes 1.98} \left[ \frac{400 - 300}{300 	imes 400} ight]$$\log(0.5) = \frac{\Delta H}{4.56} \left[ \frac{100}{120000} ight]$$-0.301 = \frac{\Delta H}{4.56} 	imes \frac{1}{1200}$$\Delta H = -0.301 	imes 4.56 	imes 1200 \approx -1646 \ cal \ mol^{-1}$.Rounding to the nearest option provided,the result is $-1652.8 \ cal$.

The equilibrium constant $(K_p)$ for the formation of ammonia from its constituent elements at $27^{\circ} C$ is $1.2 \times 10^{-4}$ and at $127^{\circ} C$ is $0.60 \times 10^{-4}$. Calculate the mean heat of formation of ammonia per mole in this temperature range. (in $cal$)

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