The addition of 3.15 g of Ba(OH)

_{2}·8H

_{2}O to a solution of 1.52 g of NH

_{4}SCN in 100 g of water in a calorimeter caused the temperature to fall by 3.1 °C. Assuming the specific heat of the solution and products is 4.20 J/g °C, calculate the approximate amount of heat absorbed by the reaction, which can be represented by the following equation:

Ba(OH)

_{2}·8H

_{2}O(s) + 2NH

_{4}SCN(aq)

Ba(SCN)

_{2}(aq) + 2NH

_{3}(aq) + 10H

_{2}O(l)

Calculate ΔH for the reaction described by the above equation.

**My answer:** q=cmΔT

=4.20 J/g°C × (3.15 + 1.52 +100)g ×3.1°C

=1362.8 J =1.4 kJ (two significant figures)

ΔH for the reaction :

We have 3.15 g × [itex]\frac{1 mol}{315.46 g} =0.00998 [/itex]mol Barium Hydroxide Octahydrate available. 1.52 × [itex]\frac{1 mol}{76.1209 g} = 0.02[/itex] mol ammonium thiocynate available.

Since 0.02 mol of NH

_{4}SCN × [itex]\frac{1 mol Ba(OH)_2\cdot 8H_2O}{2 mol NH_4SCN}=0.01 mol Ba(OH)_2 \cdot 8H_2O [/itex]is needed. Ba(OH)

_{2}·8H

_{2}O is limiting reagent.

The reaction uses 2 mol NH

_{4}SCN and the conversion factor is [itex]\frac{1.4kJ}{0.02 mol NH_4SCN}[/itex], so, we have

[tex] \Delta {H}= 2 mol \times \frac{1.4 kJ}{0.02 mol NH_4SCN}=+6700 J [/tex]

The enthalpy change for this reaction is +6700 J and thermochemical equation is

Ba(OH)

_{2}·8H

_{2}O + 2NH

_{4}SCN

Ba(SCN)

_{2} + 2NH

_{3} + 10H

_{2}O ΔH = +6.7 kJ

Is this correct?

In my opinion, it looks correct.