The enthalpy change (Delta H) for the reactio | vedclass

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(B) The reaction is $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$.Given,$\Delta H = -92.38 \ kJ$ at $298 \ K$.The relationship between enthalpy ch

Vedclass · Accepted Answer

$-87.42$

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(B) The reaction is $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$.Given,$\Delta H = -92.38 \ kJ$ at $298 \ K$.The relationship between enthalpy change and internal energy change is given by $\Delta H = \Delta U + \Delta n_g RT$.Calculate the change in the number of moles of gaseous species,$\Delta n_g = n_{p(g)} - n_{r(g)} = 2 - (1 + 3) = -2$.Substitute the values into the equation (using $R = 8.314 \ J \ K^{-1} \ mol^{-1}$ and converting $\Delta H$ to $J$):$-92.38 	imes 10^3 \ J = \Delta U + (-2) 	imes 8.314 	imes 298$.$-92380 \ J = \Delta U - 4959.15 \ J$.$\Delta U = -92380 + 4959.15 = -87420.85 \ J$.Converting back to $kJ$,$\Delta U \approx -87.42 \ kJ$.

The enthalpy change $(\Delta H)$ for the reaction,$N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$ is $-92.38 \ kJ$ at $298 \ K$. The internal energy change $\Delta U$ at $298 \ K$ is ...... $kJ$.

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