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Step-by-Step Solution
Step 1: Identify the Given Data
• The reaction given is:
NH_2CN_{(s)} + \frac{3}{2} O_2{(g)} \longrightarrow N_2{(g)} + CO_2{(g)} + H_2O_{(l)}
• The change in internal energy, \Delta U , is provided as -742.24\ \mathrm{kJ\,mol^{-1}} .
• The gas constant, R = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}} .
• Temperature, T = 298\ \mathrm{K} .
Step 2: Calculate the Change in Moles of Gas, \Delta n_g
We only count moles of gaseous species in the product and reactant sides for \Delta n_g .
Reactants (in gas phase): \frac{3}{2}\ O_2 → total moles of gas = 1.5
Products (in gas phase): N_2 + CO_2 → total = 1 + 1 = 2
\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) = 2 - 1.5 = 0.5
Step 3: Use the Relation Between \Delta H and \Delta U
The relation is given by:
\Delta H = \Delta U + \Delta n_g\,R\,T
Here:
\Delta U = -742.24\ \mathrm{kJ\,mol^{-1}}
\Delta n_g = 0.5
R = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}} = 8.314 \times 10^{-3}\ \mathrm{kJ\,mol^{-1}\,K^{-1}}
T = 298\ \mathrm{K}
Step 4: Substitute the Values and Compute \Delta H
First, compute the term \Delta n_g\,R\,T :
0.5 \times (8.314 \times 10^{-3}\ \mathrm{kJ\,mol^{-1}\,K^{-1}}) \times 298\ \mathrm{K} \\
= 0.5 \times 8.314 \times 10^{-3} \times 298 \\
= 0.5 \times 2.476 \\
= 1.238 \approx 1.24\ \mathrm{kJ\,mol^{-1}}
Next, add this term to \Delta U to find \Delta H :
\Delta H = -742.24\ \mathrm{kJ\,mol^{-1}} + 1.24\ \mathrm{kJ\,mol^{-1}} = -741\ \mathrm{kJ\,mol^{-1}}
Step 5: State the Final Answer
Thus, the enthalpy change for the reaction, rounded to the nearest integer, is -741\ \mathrm{kJ\,mol^{-1}} .
