Question

What is the change in enthalpy when 4.00 mol of sulfur trioxide decomposes to sulfur dioxide and oxygen gas? 250 2(g) + O2(g) ® 250 31 g): D Hº = 198 kJ A. -396 kJ B. 198 kJ C.-198 kJ D. 396 kJ E. 792 kJ

Answer 1

Answer: A. -396 kJ

Explanation:

The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.

$2SO_2(g)+O_2 (g)\rightarrow 2SO_3(g)$  $\Delta H^0=198kJ$

Reversing the reaction, changes the sign of $\Delta H$

$2SO_3(g)\rightarrow 2SO_2(g)+O_2 (g)$  $\Delta H^0=-198kJ$

On multiplying the reaction by 2, enthalpy gets multiplied by 2:

$4SO_3(g)\rightarrow 4SO_2(g)+2O_2 (g)$   $\Delta H=2\times -198kJ=-396kJ$

Thus the enthalpy change for the reaction $4SO_3(g)\rightarrow 4SO_2(g)+2O_2 (g)$  is -396 kJ.