Question

The heat of combustion of methane at 298 K is expressed by:

CH4(g)+2O2(g)→CO2(g)+2H2O(l);CH4(g)+2O2(g)→CO2(g)+2H2O(l);

ΔΔH=-890.2 KJ ΔEΔE of the reaction at this temperature is?

• A. less than$\Delta$H
• B. more than $\Delta$H
• C. equal to $\Delta$H
• D. not equal to $\Delta$H

Answer 1

The correct option is B more than $\Delta$H

$\Delta H=\Delta E+\Delta n_{g}RT$
$C{H}_4\left(g\right)+2O_2\left(g\right)\to C{O}_2\left(g\right)+2H_{2}O\left(l\right)$
$\Delta n_g=1-1-2$
$=-2$
$\therefore \Delta H=\Delta E-2RT$
$\therefore \Delta E=\Delta H+2RT$
$\therefore \Delta E$ is more than $\Delta H$