The addition of 3.15 g of Ba(OH)
2·8H
2O to a solution of 1.52 g of NH
4SCN 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
2O(s) + 2NH
4SCN(aq)
Ba(SCN)
2(aq) + 2NH
3(aq) + 10H
2O(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
4SCN × [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
2O is limiting reagent.
The reaction uses 2 mol NH
4SCN 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
2O + 2NH
4SCN
Ba(SCN)
2 + 2NH
3 + 10H
2O ΔH = +6.7 kJ
Is this correct?
In my opinion, it looks correct.