Answer to Question #87420 in General Chemistry for jason

Question

What is the standard enthalpy change for vaporizing 4.000 mol C5H12O(L) at 25°C?

ΔH° = ________ kJ

Answer 1

$$C_5H_{12}O(l) \rightarrow C_5H_{12}O(g)$$

$\Delta H_f^0(C_5H_{12}O(l)) = -351.6 \frac{kJ}{mol}$

$\Delta H_f^0 (C_5H_{12}O(g)) = -294.6 \frac{kJ}{mol}$

$$\Delta H_{vap} = \sum \Delta H_{products} - \sum \Delta H_{reactants}$$

$$\Delta H_{vap} = \Delta H_f^0 (C_5H_{12}O(g))-\Delta H_f^0(C_5H_{12}O(l))= -294.6 \frac{kJ}{mol} -(-351.6 \frac{kJ}{mol}) = 57 \frac{kJ}{mol}$$

$$\Delta H^0 = \Delta H_{vap} \times n = 57\frac{kJ}{mol}\times 4.000 mol = 228.0 kJ$$