Question

When one mole of sulfur burns to form $S{O}_2$, 1,300 calories are released. When one mole of sulfur burns to form $S{O}_3$, 3,600 calories are released. What is the $\Delta H$ when one mole of $S{O}_2$ is burned to form $S{O}_3$?

• A. 3,900 cal
• B. -1,950 cal
• C. 1,000 cal
• D. -500 cal
• E. -2, 300 cal

Answer 1

The correct option is E -2, 300 cal

Let's write out our information in a chemical way:
(1) $S+O_2\to S{O}_2\Delta H=-1.3kcal$
(2) $S+\left(3/2\right)O_2\to S{O}_3\Delta H=-3.6kcal$
$S{O}_2+\left(1/2\right)O_2\to S{O}_3$
To obtain the target equation, what I will do is flip the first equation, remembering to also change the sign on the enthalpy. Here are equations (1) and (2) with (1) flipped:
(1) $S{O}_2\to S+O_2\Delta H=+1.3kcal$
(2)$S+\left(3/2\right)O_2\to S{O}_3\Delta H=-3.6kcal$
When those two equations are added, an O2 will cancel out, leaving (1/2)O2 on the right-hand side. (The S also cancels.) We add the enthalpies:
+1.3 plus -3.6 = -2.3 kcal or -2300 cal