Question

A volume of 50.0 mL of 0.400 M HBr at 24.35°C is added to 50.0 mL of 0.400 M NaOH, also at 24.35°C. After the reaction, the final temperature is 27.06°C. Calculate the enthalpy change, H, per mole of HBr (in kJ) for the reaction. Heat specific capacity of the solution is 4.184 J/oC, density of the solution is 1.0 g/mL.

Answer 1

Answer:

0.567 kJ

Explanation:

In the given problem, 50.0 mL of 0.400 M HBr at 24.35°C is added to 50.0 mL of 0.400 M NaOH, also at 24.35°C. After the reaction, the final temperature is 27.06°C. To calculate the enthalpy change per mole of HBr (in kJ), we have:

The balanced chemical equation for the reaction is shown below.

$HBr_{(aq)} + NaOH_{(aq)}$⇒$NaBr_{(aq)} + H_{2}O_{(l)}$

We also need to calculate the number of moles of the reactants in the system.

For the system above, the enthalpy change is:

ΔH = heat capacity * temperature change = 4.184 J/°C * (27.06 - 24.35) °C = 11.339 J

number of moles (n) = concentration (c) *volume (v)

For HBr, n = 0.05*0.4 = 0.02 moles

For NaOH, n = 0.05*0.4 = 0.02 moles

Thus, to calculate the enthalpy change by mole, we have:

11.339 J/0.02 moles = 566.93 J

Therefore, the enthalpy change per mole of HBr (in kJ) is 0.567 kJ