Question

When 3.27 g of ammonium perchlorate (NH4ClO4) is dissolved in 102 g of water in a styrofoam calorimeter of negligible heat capacity, the temperature drops from 25.00 to 22.75 °C. Based on this observation, calculate q for the water and ΔH° for the process, assuming that the heat absorbed by the salt is negligible.
NH4ClO4(s) NH4+(aq) + ClO4- (aq)
The specific heat of water is 4.184 J °C-1 g-1.

Give the answers in kJ.

qH2O = ______ kJ

ΔH° = _______kJ

Answer 1

$$\Delta H^0 = -q (water)$$

$$q (water) = cm\Delta T$$

$$q(water) = 4.184 \frac{J}{g\times ^\circ C} \times 102 g \times (22.75 ^\circ C - 25^\circ C) = -960.228 J = -0.96 KJ$$

$$\Delta H ^0 = -q = -(-0.96 KJ) = 0.96 KJ$$

Note: usually the requiment of the task is to determine mollar enthalpy of dissolution. If so, one needs to find moles of soluteand then ratio $\frac{\Delta H^0}{n}$

$$n(NH_4ClO_4) = \frac{m}{M} = \frac{3.27 g}{117.5 \frac{g}{mol}} = 0.0278 mol$$

$$\Delta H_{dissolution} = \frac{0.96 KJ}{0.0278 mol} = 34.53 \frac{KJ}{mol}$$

Answer: $q(water) = -0.96 KJ$

$\Delta H^0 = 0.96 KJ$