A 100. mL sample of 0.200 M aqueous hydrochloric acid is added to 100. mL of 0.200 M aqueous ammonia in a calorimeter whose heat capacity (excluding any water) is 480. J/K. The following reaction occurs when the two solutions are mixed. HCl (aq) + NH3 (aq) → NH4Cl (aq) The temperature increase is 2.34°C. Calculate ΔH per mole of HCl and NH3 reacted.

Question

Adolfo Fletcher

Chemistry

7 July, 19:30

A 100. mL sample of 0.200 M aqueous hydrochloric acid is added to 100. mL of 0.200 M aqueous ammonia in a calorimeter whose heat capacity (excluding any water) is 480. J/K. The following reaction occurs when the two solutions are mixed. HCl (aq) + NH3 (aq) → NH4Cl (aq) The temperature increase is 2.34°C. Calculate ΔH per mole of HCl and NH3 reacted.

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Shirley Farley · Answer 1

-154KJ/mol

Explanation:

mole of 100ml sample of 0.2M aqueous HCl = Molarity * volume in Liter

= 0.2 * 100 / 1000 (1L = 1000 ml) = 0.02 mol and 0.02 mole of HCl solution require 0.02 mole of ammonia according to the mole ratio in the balanced equation.

Heat loss by the reaction = heat gain by calorimeter = mcΔT + 480 J/K

where m is the mass of water = 100g + 100g = 200g since mass of 100ml of water = 100g and it is in both of them and specific heat capacity of water 4.184 J/gK

heat gain by calorimeter = (4.184 * 200 + 480) * 2.34 = 3081.3 J

ΔH per mole = heat loss / number of mole = 3081.3 / 0.02 = 154065.6 = - 154KJ/mol