Question

Julie is conducting an experiment where she placed 30.0 mL of water in a calorimeter at 10.0 °C. Then, 2.5 g of A (molar mass = 48.0 g/mol), also at 10.0°C, is added to the water in the calorimeter and the temperature of the solution increases to 29.0 °C. The following reaction is produced:
a.) What is the enthalpy of the reaction (ΔH°rxn), in kJ/mol?

Answer 1

Answer on Question #55344 - Chemistry - General chemistry

Question:

Julie is conducting an experiment where she placed 30.0 mL of water in a calorimeter at 10.0 °C. Then, 2.5 g of A (molar mass = 48.0 g/mol), also at 10.0°C, is added to the water in the calorimeter and the temperature of the solution increases to 29.0 °C. The following reaction is produced:

a.) What is the enthalpy of the reaction (ΔH°rxn), in kJ/mol?

Solution:

$$n = m / M_w = 2.5 \, \text{g} / 48.0 \, \text{g mol}^{-1} = 0.052 \, \text{mol}$$ $$C_{H_2O} = (C_{10C} + C_{30C}) / 2 = (4.178 + 4.192) / 2 = 4.185 \, (\text{J g}^{-1} \, \text{K}^{-1})$$ Q = m_{H_2O} \times C_{H_2O} \times \Delta T = \rho_{H_2O} \times V_{H_2O} \times C_{H_2O} \times \Delta T = 1.00 \, \text{g mL}^{-1} \times 30.0 \, \text{mL} \times 4.185 \, \text{J g}^{-1} \, \text{K}^{-1} \times (29.0 \, ^\circ\text{C} - 10.0 \, ^\circ\text{C}) = 2385.45 \, \text{J} $$\Delta H = Q / n = Q \times M_w / m = 2385.45 \, \text{J} \times 48.0 \, \text{g mol}^{-1} / 2.5 \, \text{g} = 45800 \, \text{J/mol} = 45.8 \, \text{kJ/mol}$$

Answer 45.8 kJ/mol

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