Question

When a 3.25 g sample of solid sodium hydroxide was dissolved in a calorimeter in 100.0 g of water, the temperature rose from 23.9 °C to 32.0 °C. Calculate ∆H (in kJ/mol) for the solution process: NaOH (s) → Na+ (aq) + OH- (aq)
Use a calorimeter heat capacity of Ccal = 15.8 J/°C

Answer 1

Answer : The enthalpy change for the solution is 42.8 kJ/mol

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m\times c_2\times \Delta T]$

where,

q = heat released by the reaction

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the water

$c_1$ = specific heat of calorimeter = $15.8J/^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m$ = mass of water = 100.0 g

$\Delta T$ = change in temperature = $T_2-T_1=(32.0-23.9)=8.1^oC$

Now put all the given values in the above formula, we get:

$q=[(15.8J/^oC\times 8.1^oC)+(100.0g\times 4.18J/g^oC\times 8.1^oC)]$

$q=3513.8J=3.5138kJ$        (1 kJ = 1000 J)

Now we have to calculate the enthalpy change for the solution.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat released = 3.5138 kJ

m = mass of NaOH = 3.25 g

Molar mass of NaOH = 40 g/mole

$\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{3.25g}{40g/mole}=0.0812mole$

Now,

$\Delta H=\frac{3.5138kJ}{0.0821mole}=42.8kJ/mol$

Therefore, the enthalpy change for the solution is 42.8 kJ/mol