Question

The work done during combustion of 0.090 kg of ethane (molar mass = 30) at 300 K is:

Given: $R=8.314J{K}^{-1}mo{l}^{-1}$

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

Answer 1

Answer: (B)

$18.7$ kJ

$2C_2{H}_{6\left(g\right)}+7O_{2\left(g\right)}\to 4C{O}_{2\left(g\right)}+6H_2{O}_{\left(l\right)}$ or

$C_2{H}_6+7/2O_2\to 2C{O}_2+3H_{2}O$

Change in number of moles of gas is

$\Delta n=(2-(1+7/2\left)\right)=-2.5$

Now,

Work done, $W=-P\Delta V=\Delta nRT$

$W=-\Delta nRT=-2.5\times 8.314\times 300$

$W=-(-2.5\times 8.314\times 300)=6235.5$ J

For one mole combustion, we get work done $=6235.5$ J

So, for $0.090$ kg $=0.090\times 1000/30$ moles $=3$ moles

For 3 moles the work done $=3\times 6235.5$ J $=18706.5=18.7$ kJ