Question

When 25.0 mL of 0.500 M H2SO4 is added to 25.00 mK of 1.00 M KOH in a coffee-cup calorimeter at 23.50◦ C, the temperature rises to 30.17◦ C. Calculate DHrxn for this reaction. (Assume that the density of water of and specific heat capacity of the solution are the same as for pure water). (Answer: –55.8 kJ/mol KOH, –112 kJ/mol H2SO4)

Answer 1

Answer:

$\Delta H_rxn=-55.8\frac{kJ}{mol KOH}$

$\Delta H_rxn=-111.6\frac{kJ}{mol H2SO4}$

Explanation:

The reaction: $H_2SO_4 (aq) + 2 KOH (aq) \longrightarrow K_2SO_4 (aq) + H_2O (l)$

Total volume: V=50 mL

The heat generated:

$Q=Cp_{water}*\Delta T*V*\delta_{water}$

$Q=4.184 \frac{J}{g*C}*(30.17 C -23.5 C)*50 mL*\frac{1 g}{mL}}$

$Q=1395.4 J$

Moles of KOH: $n_K=\frac{1 mol}{1000 mL}*25 mL=0.025 mol$

Moles of H2SO4: $n_S=\frac{0.5 mol}{1000 mL}*25 mL=0.0125 mol$

Heat of reaction:

$\Delta H_{rxn}=\frac{-1395.4 J}{0.025 mol KOH}*\frac{1 kJ}{1000 J}$

$\Delta H_{rxn}=-55.8\frac{kJ}{mol KOH}$

$\Delta H_{rxn}=\frac{-1395.4 J}{0.0125 mol H2SO4}*\frac{1 kJ}{1000 J}$

$\Delta H_{rxn}=-111.6\frac{kJ}{mol H2SO4}$

Note: the negative sign is beacuse it is an exothermical reaction.