Question

Calculate the enthalpy change in the burning of 4.000 mol gaseous 1-pentanol to form gaseous products at 25°C.

ΔH° = _______ kJ

Answer 1

$$2C_5H_{12}O(g) + 15 O_2(g)\rightarrow 10CO_2(g) + 12H_2O(g)$$

$\Delta H_f^0 (C_2H_{12}O(g)) = -294.6\frac{kJ}{mol}$

$\Delta H_f^0 (CO_2(g)) = -393.5 \frac{kJ}{mol}$

$\Delta H_f^0 (H_2O(g)) = -241.8 \frac{kJ}{mol}$

$$\Delta H_{rxn}^0 = \sum \Delta H_{products} - \sum \Delta H_{rectants}$$

$$\Delta H_{rxn}^0 =[10\times \Delta H_f^0(CO_2(g))+12\times \Delta H_f^0(H_2O(g))] - [2\times \Delta H_f^0 (C_2H_{12}O(g))+ 15\times \Delta H_f^0(O_2(g))] = [10\times(-393.5)+12\times(-241.8)]- [2\times(-294.6)+15\times 0] = 6247.4 \frac{kJ}{mol}$$

$$\Delta H^0 = \Delta H_{rxn}^0 \times n = -6247.4\frac{kJ}{mol}\times 4.000 mol = -24989.6 kJ$$

As $\Delta H^0<0$ , then $Q>0$ , reaction is exothermic

Answer: -24989.6 kJ, exothermic