Question

Sulphur trioxide is prepared by the following two reactions.
$S_8\left(s\right)+8O_2\left(g\right)\to 8S{O}_2\left(g\right);\Delta H=-2374.kJ$
$2S{O}_2\left(g\right)+O_2\left(g\right)\to 2S{O}_3\left(g\right);\Delta H=-198.kJ$
Calculate the net change in enthalpy for the formation of one mole of sulfur trioxide from sulfur and oxygen.

• A. $-199$ kJ
• B. $-198$ kJ
• C. $396$ kJ
• D. $-296.75$ kJ

Answer 1

The correct option is C $396$ kJ

$S_8\left(s\right)+8O_2\left(g\right)\to 8S{O}_2\left(g\right);\Delta H=-2374.kJ$
$2S{O}_2\left(g\right)+O_2\left(g\right)\to 2S{O}_3\left(g\right);\Delta H=-198.kJ$
Divide the first equation by 8 to show 1 $S{O}_2$ being produced.
Divide the first equation by 2 to show 1 $S{O}_3$ being produced.
Add the two equations to get the overall equation.
$1/8S_8+O_2\to S{O}_2;\Delta H=-296.75kJ$
$S{O}_2\left(g\right)+1/2O_2\left(g\right)\to S{O}_3\left(g\right);\Delta H=-199.kJ$

$1/8S_8+3/2O2\to S{O}_3;\Delta H=-396kJ$