Question

The ΔH for the solution process when solid sodium hydroxide dissolves in water is 44.4 kJ/mol. When a 6.21-g sample of NaOH dissolves in 250.0 g of water in a coffee-cup calorimeter, the temperature increases from 23.0 °C to ________°C. Assume that the solution has the same specific heat as liquid water, i.e., 4.18 J/g-K.

Answer 1

Answer: The final temperature of the solution is 29.6°C

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of NaOH = 6.21 g

Molar mass of NaOH = 40 g/mol

Putting values in above equation, we get:

$\text{Moles of NaOH}=\frac{6.21g}{40g/mol}=0.155mol$

To calculate the enthalpy change of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat absorbed = ?

n = number of moles = 0.155 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction = 44.4 kJ/mol = 44400 J/mol   (Conversion factor:  1 kJ = 1000 J)

Putting values in above equation, we get:

$44400J/mo=\frac{q}{0.155mol}\\\\q=(44400J/mol\times 0.155mol)=6882J$

To calculate the heat absorbed by the calorimeter, we use the equation:

$q=mc\Delta T$

where,

q = heat absorbed = 6882 J

m = mass of water = 250 g

c = heat capacity of solution = 4.18 J/g.K = 4.18 J/g°C

$\Delta T$ = change in temperature = $T_2-T_1=(T_2-23)^oC$

Putting values in above equation, we get:

$6882J=250g\times 4.18J/g^oC\times (T_2-23)\\\\T_2=29.6^oC$

Hence, the final temperature of the solution is 29.6°C