Question

200 mL of a 2.5 mole/L pf a solution of phosphoric acid reacts with 300 mL of a 3.0 mole/L solution of a potassium hydroxide. The reaction occurs in a glass calorimeter of mass 1000 g. Calculate the energy released and the temperature change which the calorimeter will undergo.

Answer 1

Question #58811, Chemistry, Other

200 mL of a 2.5 mol/L solution of phosphoric acid reacts with 300 mL of a 3.0 mol/L solution of a potassium hydroxide. The reaction occurs in a glass calorimeter of mass 1000 g. Calculate the energy released and the temperature change which the calorimeter will undergo.

Answer:

$$\mathrm{H_3PO_4} + 3\mathrm{KOH} = \mathrm{K_3PO_4} + 3\mathrm{H_2O} + 173.2\ \mathrm{kJ/mol}$$ $$\mathrm{C_M = v/V} \quad \mathrm{v = C_M \cdot V}$$ $$\begin{array}{l} v \left(\mathrm{H_3PO_4}\right) = 2.5 \cdot 0.2 = 0.5\ \mathrm{mol} \\ v \left(\mathrm{KOH}\right) = 3.0 \cdot 0.3 = 0.9\ \mathrm{mol} \\ \end{array}$$

However, according to the equation, the required amount of KOH is: $v \left(\mathrm{KOH}\right)_{\text{need}} = 0.5 \cdot 3 = 1.5\ \mathrm{mol}$

The real amount of KOH is less. That is why, KOH is a limiting reagent for the reaction. The amount of $\mathrm{K_3PO_4}$ formed must be calculated from this value.

$$\begin{array}{l} v \left(\mathrm{K_3PO_4}\right) = 0.9 / 3 = 0.3\ \mathrm{mol} \\ Q_{\text{reaction}} = 173.2 \cdot 0.3 = 51.96\ \mathrm{kJ} = 51960\ \mathrm{J} \\ Q = c \cdot m \cdot \Delta T \\ \Delta T = Q / (c \cdot m) \\ c \text{ (glass)} = 0.840\ \mathrm{J/g \cdot ^\circ C} \\ \Delta T = 51960 / (0.840 \cdot 1000) = 61.857\ \mathrm{^\circ C} \\ \end{array}$$

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