The enthalpy change left( Delta H right) for | vedclass

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(B) The reaction is $N_{2(g)} + 3H_{2(g)} 	o 2NH_{3(g)}$.Given: $\Delta H = -92.38 \ kJ$,$T = 298 \ K$,$R = 8.314 	imes 10^{-3} \ kJ \ K^{-1} \ mol^

Vedclass · Accepted Answer

$-87.42$

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(B) The reaction is $N_{2(g)} + 3H_{2(g)} 	o 2NH_{3(g)}$.Given: $\Delta H = -92.38 \ kJ$,$T = 298 \ K$,$R = 8.314 	imes 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$.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,$\Delta n_g = n_{products} - n_{reactants} = 2 - (1 + 3) = -2$.Substitute the values into the equation: $-92.38 = \Delta U + (-2) 	imes (8.314 	imes 10^{-3}) 	imes 298$.$-92.38 = \Delta U - 4.959$.$\Delta U = -92.38 + 4.959 = -87.421 \ kJ \approx -87.42 \ kJ$.

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

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