## Chemistry: The Molecular Science (5th Edition)

(a) $-2.1 \times 10^2 kJ$ (b) $-33kJ$
(a) 1. Divide the amount (in mols) by the coefficient of $H_2O(l)$: $\frac{34.2}{ 1} = 34.2 mol - reaction$ 2. Find the necessary heat: ** Since the water freezes, the reaction is reversed, so the $\Delta_rH^{\circ}$ is negative. $34.2mol-reaction \times -6.0 kJ/mol-reaction = -2.1 \times 10^2 kJ$ ---------------------- (b) Molar Mass ($H_2O$): 1.008* 2 + 16* 1 = 18.016g/mol 1. Calculate the amount in moles: $100g \times \frac{1 mol (H_2O)}{ 18.016} = 5.551 mol$ 2. Divide the amount by the coefficient of $H_2O(l)$: $\frac{5.551}{ 1} = 5.551 mol - reaction$ 3. Find the necessary heat: ** Since the water freezes, the reaction is reversed, so the $\Delta_rH^{\circ}$ is negative. $5.551mol-reaction \times -6.0 kJ/mol-reaction = -33kJ$