Question

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

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 conditions ($T = 298$ K and $R=8.314$ J/(mol·K)), $\Delta n_{gas}=(1 + 3)-2=2$. First, convert $\Delta n_{gas}RT$ to kJ: $\Delta n_{gas}RT=2\times8.314$ J/(mol·K)$\times298$ K $=4955.144$ J $=4.955$ kJ. Then, $\Delta U^{\circ}=\Delta H^{\circ}-\Delta n_{gas}RT=91.8$ kJ - 4.955 kJ $=86.8$ kJ.

Answer:

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