Answer to Question #173325 in Organic Chemistry for vinay tiwari

Question #173325

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.


1
Expert's answer
2021-03-22T06:33:44-0400

"\\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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