Question

Given $R=8.314J{K}^{-1}{mol}^{-1}$, the work done during combustion of $0.090kg$ of ethane (molar mass $=30$) at $300K$ is:

• A. $-18.7kJ$
• B. $18.7kJ$
• C. $6.234kJ$
• D. $-6.234kJ$

Answer 1

Answer: (B)

$18.7kJ$

$C_2{H}_{6\left(g\right)}+\frac{7}{2}{O}_{2\left(g\right)}⟶{CO}_{2\left(g\right)}+H_2{O}_{\left(l\right)}$

${\Delta n}_g=2-(1+\frac{7}{2})=\frac{-5}{2}=-2.5$

$\therefore$ work done for 1 mole of $C_2{H}_6=-{\Delta n}_g\times R\times T$

$\therefore$ work done for $\frac{0.090\times {10}^3}{30}$ moles of $C_2{H}_6$ $=\frac{0.090\times {10}^3}{30}\times \left(2.5\right)\times 8.314\times 300$

$=18.7KJ$