Question

When solid sodium hydroxide dissolves in water, the ΔH for the solution process is −44.4 kJ/mol. If a 13.9 g sample of NaOH dissolves in 250.0 g of water in a coffee-cup calorimeter initially at 23.0 °C. What is the final temperature of the solution? Assume that the solution has the same specific heat as liquid water, i.e., 4.18 J/g·K.
a.24.0º C
b.40.2º C
c.37.8º C
d.35.2º C
e.37.0º C

Answer 1

Answer: The final temperature is 8.13°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 = 13.9 g

Molar mass of NaOH = 40 g/mol

Putting values in above equation, we get:

$\text{Moles of NaOH}=\frac{13.9g}{40g/mol}=0.35mol$

• To calculate the heat released in the reaction, we use the equation:

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

where,

$q$ = amount of heat released

n = number of moles = 0.35 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction = -44.4 kJ/mol

Putting values in above equation, we get:

$-44.4kJ/mol=\frac{q}{0.35mol}\\\\q=(-44.4kJ/mol\times 0.35mol)=-15.54kJ$

• To calculate the final temperature, we use the equation:

$q=mc\Delta T=mc(T_2-T_1)$

where,

q = heat released = -15.54 kJ = -15540 J    (Conversion factor: 1 kJ = 1000 J)

m = mass of water = 250.0 g

c = specific heat capacity of water = 4.18 J/g.K

$T_1$ = initial temperature = $23^oC=[23+273]=296K$

$T_2$ = final temperature = ?

Putting values in above equation, we get:

$-15540J=250g\times 4.18J/g.K\times (T_2-296)\\\\T_2=281.13K$

Converting the temperature from kelvins to degree Celsius, by using the conversion factor:

$T(K)=T(^oC)+273$

$281.13=T(^oC)+273\\T(^oC)=8.13^oC$

Hence, the final temperature is 8.13°C