Question

An exothermic reaction with a 1-to-1 stoichiometric ratio and known δ h r x n = − 3.45 k j / m o l is performed in a constant pressure calorimeter. if after adding 0.469 mol of limiting reagent, the temperature of the calorimeter is seen to increase from 293.15 k to 299.27 k. what is the heat capacity of the calorimeter ( c c a l i n j k − 1 ) ?

Answer 1

The heat capacity of the calorimeter (Cc) is calculated using the heat absorbed by it during the exothermic reaction and the change in temperature. It is found to be 264.5 J/K.

Step-by-step explanation:

To determine the heat capacity of the calorimeter (Cc), we use the formula q = m × ΔT × C, where q is the heat absorbed by the calorimeter, m is the amount of substance in moles, ΔT is the change in temperature, and C is the specific heat capacity. Since the reaction is exothermic (δHrxn = -3.45 kJ/mol), the heat released by the reaction is the negative of the heat absorbed by the calorimeter. Therefore, q for the calorimeter is positive when it absorbs the released heat. Given that 0.469 moles of the reagent were used and the temperature increased by 6.12 K (299.27 K - 293.15 K), we can calculate the heat absorbed using q = m × δHrxn, yielding q = 0.469 moles × -3.45 kJ/mol = -1.61805 kJ (since the reaction is exothermic, the sign of δHrxn must be considered).

The heat absorbed by the calorimeter is equal to the heat released by the reaction but with an opposite sign, so qcal = 1.61805 kJ. With the temperature change known, we now solve for the calorimeter’s heat capacity using q = Cc × ΔT, resulting in Cc = q / ΔT = 1.61805 kJ / 6.12 K which approximates to 0.2645 kJ/K or 264.5 J/K as the heat capacity of the calorimeter.