Question

question 4 (1 point)
find δu° and δh°reaction for the following reaction:
2 nh3(g) → n2(g) + 3 h2(g)
the standard enthalpy of formation of nh3 is -45.9 kj/mol
δh°=91.8, δu°= 86.8 kj
δh°=0.991, δu°= 0.086 kj
δh°=45.9, δu°= -45.9 kj
δh°=-45.9, δu°= +45.9 kj
view hint for question 4

Explanation:

Step1: Calculate $\Delta H^{\circ}$

The standard - enthalpy of formation of elements in their standard states is 0. For the reaction $2NH_3(g)
ightarrow N_2(g)+3H_2(g)$, $\Delta H^{\circ}=\sum n_p\Delta H_f^{\circ}(products)-\sum n_r\Delta H_f^{\circ}(reactants)$. Here, $\Delta H_f^{\circ}(N_2) = 0$, $\Delta H_f^{\circ}(H_2)=0$ and $\Delta H_f^{\circ}(NH_3)=- 45.9\ kJ/mol$. So, $\Delta H^{\circ}=0 + 0-2\times(-45.9\ kJ/mol)=91.8\ kJ$.

Step2: Calculate $\Delta U^{\circ}$

The relationship between $\Delta H^{\circ}$ and $\Delta U^{\circ}$ is $\Delta H^{\circ}=\Delta U^{\circ}+\Delta n_{gas}RT$. At standard - state conditions ($T = 298\ K$) and for the reaction $2NH_3(g)
ightarrow N_2(g)+3H_2(g)$, $\Delta n_{gas}=(1 + 3)-2=2$. Using $R = 8.314\ J/(mol\cdot K)=8.314\times10^{-3}\ kJ/(mol\cdot K)$ and $T = 298\ K$, $\Delta n_{gas}RT=2\times8.314\times10^{-3}\ kJ/(mol\cdot K)\times298\ K\approx4.95\ kJ$. Then $\Delta U^{\circ}=\Delta H^{\circ}-\Delta n_{gas}RT=91.8\ kJ - 4.95\ kJ\approx86.8\ kJ$.

Answer:

$\Delta H^{\circ}=91.8,\Delta U^{\circ}=86.8\ kJ$