Question

What is the final temperature of the solution formed when 0.75 grams of KMnO4 is added to 85.5 grams of water at 13.2 degrees Celsius in a calorimeter

Answer 1

$12.7^{0} \mathrm{C}$ is the final temperature of the given solution.

Explanation:

When heat is absorbed during reaction enthalpy is positive. To calculate amount of heat absorbed by solution:

$\mathbf{Q}=\Delta \mathbf{H} \cdot \mathbf{n}\left(\mathbf{K} \mathbf{M} \mathbf{n} \mathbf{O}_{4}\right)$

To calculate no of moles of $\mathrm{KMnO}_{4}$ molar mass is 158.034 g/mol:

$\mathbf{n}\left(\boldsymbol{K} \boldsymbol{M} \boldsymbol{n} \boldsymbol{O}_{4}\right)=\frac{\text { Mass of } K M n \boldsymbol{O}_{4}}{\text { Molar Mass of } K M n O_{4}}$

$\mathbf{n}\left(\boldsymbol{K} M \boldsymbol{n} \boldsymbol{O}_{4}\right)=\frac{0.75 \boldsymbol{g}}{158.034 g m o l^{-1}}$

$\mathbf{n}\left(\boldsymbol{K} M \boldsymbol{n} \boldsymbol{O}_{4}\right)=\mathbf{0 . 0 0 4 7 5} \mathbf{m o l}$

Hence quantity of heat can be calculated as given:

$\mathrm{KMnO}_{4}(\mathrm{s}) \longrightarrow \mathrm{K}^{+}(\mathrm{aq})+\mathrm{MnO}_{4}(\mathrm{aq}) ; \Delta \mathbf{H}=+42.1 \mathrm{kJ} / \mathrm{mol}$

As considered that heat capacity of pure water is same as of solution (means to heat 1 gm of water by 1 degree Celsius the energy required) = $4.186 \mathrm{J} / \mathrm{g}^{0} \mathrm{C}$

Now it is considered that due to dissolvation, no heat loss occurs and final temperature can be calculated by using specific heat capacity:

$T_{2}=\frac{-Q}{c m_{s o l u t i o n}}+T_{1}$

$T_{2}=\frac{-\left(0.20 \times 10^{3} J\right)}{\left(4.186 J / g^{0} C\right) \times(86.6 g)}+13.2$

$T_{2}=\frac{-\left(0.20 \times 10^{3} J\right)}{362.507 J /^{\circ} C}+13.2$

$T_{2}=-0.551+13.2$

$T_{2}=12.7^{\circ} \mathrm{C}$

The final temperature of the solution is $12.7^{\circ} \mathrm{C}$.