Question

When 21.45 g of KNO3 was dissolved in water in a calorimeter, the temperature fell from 25.00°C to 14.14 °C. If the heat capacity is 0.505KJ/°C, what will be the delta H for the solution process.

Answer 1

25.9 kJ/mol. (3 sig. fig. as in the heat capacity.)

Explanation

The process:

$\text{KNO}_3\;(s) \to \text{KNO}_3\;(aq)$.

How many moles of this process?

Relative atomic mass from a modern periodic table:

• K: 39.098;
• N: 14.007;
• O: 15.999.

Molar mass of $\text{KNO}_3$:

$M(\text{KNO}_3) = 39.098 + 14.007 + 3\times 15.999 = 101.102\;\text{g}\cdot\text{mol}^{-1}$.

Number of moles of the process = Number of moles of $\text{KNO}_3$ dissolved:

$\displaystyle n = \frac{m}{M} = \frac{21.45}{101.102} = 0.212162\;\text{mol}$.

What's the enthalpy change of this process?

$Q = C\cdot \Delta T = 0.505 \times (25.00 - 14.14) = 5.4843\;\text{kJ}$ for $0.212162\;\text{mol}$. By convention, the enthalpy change $\Delta H$ measures the energy change for each mole of a process.

$\displaystyle \Delta H = \frac{Q}{n} = \frac{5.4843\text{kJ}}{0.212162\;\text{mol}} = 25.8\;\text{kJ}\cdot\text{mol}^{-1}$.

The heat capacity is the least accurate number in these calculation. It comes with three significant figures. As a result, round the final result to three significant figures. However, make sure you keep at least one additional figure to minimize the risk of rounding errors during the calculation.