Consider the reaction: N_2(g) + 3H_2(g) to 2N | vedclass

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(C) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + \De

Vedclass · Accepted Answer

$\Delta H < \Delta U$

Vedclass · Answer

(C) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$. Here,$\Delta n_g$ is the change in the number of moles of gaseous products and reactants. For the reaction $N_2(g) + 3H_2(g) 	o 2NH_3(g)$,the number of moles of gaseous products is $n_p = 2$ and the number of moles of gaseous reactants is $n_r = 1 + 3 = 4$. Therefore,$\Delta n_g = n_p - n_r = 2 - 4 = -2$. Substituting this into the equation,we get $\Delta H = \Delta U - 2RT$. Since $R$ and $T$ are positive values,it follows that $\Delta H < \Delta U$.

Consider the reaction: $N_2(g) + 3H_2(g) \to 2NH_3(g)$ carried out at constant temperature and pressure. If $\Delta H$ and $\Delta U$ are the enthalpy and internal energy changes for the reaction,which of the following expressions is true?

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