A100. 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.

-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

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