Question

1. calculate the value of $delta h_{rxn}^{0}$ (in kj) for the following reaction:

$2h_{2}s(g)+3o_{2}(g)\to2h_{2}o(l)+2so_{2}(g)$

a. -561.8 kj
b. -1124 kj
c. -1143 kj
d. -1204 kj
e. -732.3 kj

1. estimate $delta h^{0}$ for the following reaction from the bond energies given.

Explanation:

Step1: Use the formula for calculating $\Delta H^{\circ}_{rxn}$

$$\Delta H^{\circ}_{rxn}=\sum n\Delta H^{\circ}_{f}(\text{products})-\sum n\Delta H^{\circ}_{f}(\text{reactants})$$

Step2: Calculate the sum of $\Delta H^{\circ}_{f}$ for products

For products: $2$ moles of $H_{2}O(l)$ and $2$ moles of $SO_{2}(g)$
$$\sum n\Delta H^{\circ}_{f}(\text{products})=(2\times(- 285.8)+2\times(-296.4))\text{kJ}$$
$$=( - 571.6-592.8)\text{kJ}=-1164.4\text{kJ}$$

Step3: Calculate the sum of $\Delta H^{\circ}_{f}$ for reactants

For reactants: $2$ moles of $H_{2}S(g)$ and $3$ moles of $O_{2}(g)$ (since $\Delta H^{\circ}_{f}(O_{2}(g)) = 0$)
$$\sum n\Delta H^{\circ}_{f}(\text{reactants})=(2\times(-20.15)+3\times0)\text{kJ}=-40.3\text{kJ}$$

Step4: Calculate $\Delta H^{\circ}_{rxn}$

$$\Delta H^{\circ}_{rxn}=-1164.4-(-40.3)\text{kJ}$$
$$=-1164.4 + 40.3\text{kJ}=-1124.1\text{kJ}\approx - 1124\text{kJ}$$

Answer:

B. -1124 kJ