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.
"\\frac12N_{2(g)}+ \\frac32H_{2(g)} \\to NH_{3(g)}"
"\u2206C_{p(reaction)} = C_{p(NH_3)} - \\frac32C_{p(H_2)} - \\frac12C_{p(N_2)}"
"\u2206C_{p(reaction)} = 35.06 -\\frac32(28.82) - \\frac12(29.12)"
"\u2206C_{p(reaction)} = -22.73\\ J\/kg.K"
"\\begin{aligned}\n\u2206H_{373} &= \u2206H_{273} + nC_p\u2206T\\\\\n&= -46.11 + (1\u00d7-22.73 \u00d710^{-3}\u00d7 75)\\\\ \n&= -46.11 + (-1.705)\\\\\n&=-47.816\\ kJ\/mol\n\n\\end{aligned}"
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