Question

The enthalpy of formation of methane at constant pressure and $300K$ is $-75.83KJ$. What will be the heat of formation at constant volume? $(R=8.3J{K}^{-1}mo{l}^{-1}$)

• A. -73.34 kJ
• B. 73.34 kJ
• C. 72.34 kJ
• D. -72.34 kJ

Answer 1

The correct option is A

-73.34 kJ

The equation for the formation of methane is
$C\left(s\right)+2H_2\left(g\right)\to C{H}_4\left(g\right);△H=-75.83kJ$
$△$ n = 1 - 2 = -1 (only gaseous components are taken into account)
Given $△H=-75.83KJ,R=8.3\times {10}^{-3}$ $kJ{K}^{-1}mo{l}^{-1},T=300K$
The enthalpy change can be given by the equation:
$△H=△U+△nRT$
$△U=△H-△nRT$
$△U=-75.83-(-1\times 8.3\times {10}^{-3}\times 300)$
$△U=-75.83+2.49$
$△U=-73.34kJ$