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Thermodynamics

# Find the value of logKP for the reaction  $N_2(g) + 3H_2(g) \leftrightharpoons 2NH_3(g)$  The standard enthalpy of formation of $NH_3 \: is \: -46\: kJ$. The standard entropies of $N_2(g),\: H_2(g)\: and\: NH_3(g) \: are\: 191, 130\: and\: 192\: J/Kmol$ respectively. $(R = 8.3 \: J/Kmol)$

$\begin {array} {1 1} (A)\;2.226 & \quad (B)\;222.6 \\ (C)\;-2.226 & \quad (D)\;-222.6 \end {array}$

$\Delta H^{\circ}$ for the given reaction is $-46\: kJ$
$\Delta S^{\circ} = 2 \times S^{\circ}(NH_3) - 3 \times S^{\circ}(H_2) – S^{\circ}(N_2)$
$= 2(192) – 3(130) – 191 = -197\: J/K$
$\Delta G^{\circ} = \Delta H^{\circ} – T \Delta S^{\circ} = -46 – \bigg(\large\frac{-197}{1000} \bigg) $$(298) = -12.706\: kJ Now, \Delta G^{\circ} = -2.303\: RT \: logK_P \Rightarrow log\: K_P =\large\frac{ \Delta G^{\circ}}{-2.303RT }$$= \large\frac{(-12.706 \times 1000)}{(-2.303 \times 8.314 \times 298)}$$= 2.226$
Ans : (A)