Question

The work done during combustion of $9 \times 10^{-2} \, Kg$ of ethane, $C_2H_6 \, (g)$ at $300 \,K$ is
(Given $R = 8.314 \, J \, deg^{-1} \, mol^{-1} $ , atomic mas $C = 12 , H = 1$)

• A. $6.236 \,kJ$
• B. $-6.236\, kJ$
• C. $18.71 \,kJ$
• D. $- 18.71 \,kJ$

Answer 1

Answer: (C)

Work done in a chemical reaction,
$W =\Delta n _{ g } RT$
where, $\Delta n _{ g }$ = number of moles of gaseous products - number of moles of gaseous reactants.
Combustion of ethane is-
$C _{2} H _{6}( g )+\frac{7}{2} O _{2}( g ) \rightarrow 2 CO _{2}( g )+3 H _{2} O ( l )$
$\therefore n _{ g }=2-4.5=-2.5$
$W =(2.5 \,mol )\left(8.314 \,JK ^{-1} \,mol ^{-1}\right) \times 300\, K$
$=6235.5 \,kJ =6.2355 \,J$
Now , $1$ mole of $C _{2} H _{6}=30 \,gm$ of $C _{2} H _{6}=6.2355 \,kJ$
Work done for combustion of $30\, gm$ of $C _{2} H _{6}=6.2355\, kJ$
$\therefore $ Work done for combustion of $90 \,gm$ of $C _{2} H _{6}$
$=\frac{6.2355 \times 90}{30}$
$=18.7065\, kJ =18.71 \,kJ$