Consider the reaction at 298 K:H_{2(g)} + Cl | vedclass

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(B) The relationship between enthalpy change $(\Delta H^o)$ and internal energy change $(\Delta U^o)$ is given by the equation: $\Delta H^o = \Delta U

Vedclass · Accepted Answer

$-185$

Vedclass · Answer

(B) The relationship between enthalpy change $(\Delta H^o)$ and internal energy change $(\Delta U^o)$ is given by the equation: $\Delta H^o = \Delta U^o + \Delta n_g RT$.First,calculate the change in the number of moles of gaseous species $(\Delta n_g)$:$\Delta n_g = \sum n_{g, 	ext{products}} - \sum n_{g, 	ext{reactants}}$$\Delta n_g = 2 - (1 + 1) = 2 - 2 = 0$.Since $\Delta n_g = 0$,the equation becomes:$\Delta H^o = \Delta U^o + (0)RT$$\Delta H^o = \Delta U^o$.Given $\Delta H^o = -185 \ kJ$,therefore $\Delta U^o = -185 \ kJ$.

Consider the reaction at $298 \ K$:
$H_{2(g)} + Cl_{2(g)} \to 2HCl_{(g)}, \Delta H^o = -185 \ kJ$
What is $\Delta U^o$ for this reaction? .....$kJ$

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Consider the reaction at $298 \ K$:$H_{2(g)} + Cl_{2(g)} \to 2HCl_{(g)}, \Delta H^o = -185 \ kJ$What is $\Delta U^o$ for this reaction? .....$kJ$