Question

The enthalpy of formation of NH3(g) as per the following reaction is -46.11 kJ/mol

at 298 K.

Calculate the value of enthalpy of formation of NH3(g) at 100 oC. The p C

values are given as: N2(g) = 29.12; H2(g) = 28.82; NH3(g) = 35.06. Assume the

given heat capacity values to be temperature independent in the range.

Answer 1

$\frac12N_{2(g)}+ \frac32H_{2(g)} \to NH_{3(g)}$

$∆C_{p(reaction)} = C_{p(NH_3)} - \frac32C_{p(H_2)} - \frac12C_{p(N_2)}$

$∆C_{p(reaction)} = 35.06 -\frac32(28.82) - \frac12(29.12)$

$∆C_{p(reaction)} = -22.73\ J/kg.K$

$\begin{aligned} ∆H_{373} &= ∆H_{273} + nC_p∆T\\ &= -46.11 + (1×-22.73 ×10^{-3}× 75)\\ &= -46.11 + (-1.705)\\ &=-47.816\ kJ/mol \end{aligned}$