Question

When one mole of sulfur burns to ${SO}_2$ 1300 calories are released. When one mole of sulfur burns to ${SO}_3$, 3600 calories are released. Determine the $\Delta H$ when one mole of ${SO}_2$ is burned to form ${SO}_3$?

• A. 3900 cal
• B. -1950 cal
• C. 1000 cal
• D. -500 cal
• E. -2300 cal

Answer 1

The correct option is E -2300 cal

$S+O_2\to S{O}_2△H=-1300calorie\left(1\right)S+\frac{3}{2}{O}_2\to S{O}_3△H=-3600calorie\left(2\right)S{O}_2+\frac{1}{2}{O}_2\to S{O}_3$

After reversing the $1st$ equation, we get

$S{O}_2\to S+O_2△H=1300calorie\left(2\right)\Rightarrow S+\frac{3}{2}{O}_2\to S{O}_3△H=-3600calorie$

Adding the above two equations, we get

$S{O}_2+S+\frac{3}{2}{O}_2\to S+O_2+S{O}_3△H=(1300-3600)calorieS{O}_2+\frac{1}{2}{O}_2\to S{O}_3△H=-2300calorie$