Question

The standard heat of formation of $C{H}_4\left(g\right),C{O}_2\left(g\right)and{H}_{2}O\left(g\right)$ are -76.2, -398.8 and -241.6 $kJmo{l}^{-1}$ respectively. Calculate the amount of heat evolved by burning $1m^3$ of methane measured under normal conditions.

• A. 35973.2 kJ
• B. 35972.2 kJ
• C. -35973.2 kJ
• D. 35972.2 KJ

Answer 1

The correct option is C

-35973.2 kJ

The required equation for combustion of methane is:

$C{H}_4+2O_2\to C{O}_2+2H_{2}O;△H=?$

The enthalpy change is given by:

$△$ H = $△H_{f\left(products\right)}-△H_{f\left(reactants\right)}$

$=△H_{f\left(C{O}_2\right)}+2\times △H_{f\left(H_{2}O\right)}-△H_{f\left(C{H}_4\right)}-2△H_{f\left(O_2\right)}$

$△$ H = -398.8 - 2 × 241.6 - (-76.2)

$△$ H = -805.8 kJ mol-1

Now we need to calculate for $1m^3$

1 mole = 22.4 liters

1 $m^3$ = 1000 liters

= -805.8/22.4 × 1000

= -35973.2 kJ