Question

A sample of 14.5 g of sodium bicarbonate (NaHCO3) was dissolved in 100 ml of water in a coffee-cup calorimeter with no lid by the following reaction
NaHCO3 (s) → Na+ (aq) + HCO3- (aq)

If the temperature of the water and the calorimeter (heat capacity of calorimeter = 150 J/oC) decreases from 25.00oC to 19.86oC, (1) what is the ΔH of the reaction?

Answer 1

Answer:

The ΔH of the reaction is + 12.45 KJ/mol

Explanation:

Mass of water= 100ml = 100g. (You should always assume 1cm3 of water as 1g)

heat capacity of water = 4.18 Jk-1 Mol-1

Change in temperature = (19.86 - 25.00) = -5.14 K (This is an endothermic reaction because of the fall in temperature)

Molar mass of NaHCO3 = 84 g/mol

Mole of NaHCO3 = 14.5 / 84 = 0.173 mol

Step 1 : Calculate the heat energy (Q) lost by the water.

            Q = M x C x ΔT

            Q = -100 x 4.18 x (-5.14)

            Q = 2148.5 joules

            Q = 2.1485 K J

Step 2: Calculating the ΔH of the reaction?

          ΔH = Q / number of moles of NaHCO3

          ΔH = 2.1485 / 0.173

          ΔH = 12.42 KJ/mol