Question

The enthalpy of solution of sodium hydroxide is –44.4 kJ/mol. When a 13.9-g sample of NaOH dissolves in 250.0 g of water 23.0 °C in a coffee-cup calorimeter, what is the final temperature of the solution assuming no heat is lost to the surroundings. The solution has the same specific heat of 4.184 J/g-K.

Answer 1

Answer:

The final temperature of the NaOH solution: T₂ = 37 °C  

Explanation:

Given: The enthalpy of NaOH solution: $\Delta H_{sol}$ =  – 44.4 kJ/mol,

The specific heat capacity of solution: c = 4.184 J/g.K

Given mass of solute (NaOH): w = 13.9 g, Molar mass of NaOH: m = 40 g/mol

Mass of the solvent: W = 250 g

Initial temperature of the solution: T₁ = 23 °C = 23 + 273 = 296 K    (∵ 0°C = 273 K)

Final temperature of the solution: T₂ = ?

The number of moles of NaOH = $\frac{given\: mass}{molar\: mass} = \frac{13.9\, g}{40\, g/mol} = 0.3475 mol$

So the amount of heat released (ΔH) by dissolution of 13.9 g NaOH:

$\Delta H = \Delta H_{sol} \times n$

$\Rightarrow \Delta H = (- 44.4 kJ/mol) \times (0.3475 mol) = 15.429 kJ$

$\Rightarrow \Delta H = 15.429 kJ = 15429 J$

(∵ 1 kJ = 1000J )

Since the specific heat capacity is given by the equation:

$c = \frac{\Delta H}{M\times \Delta T}$

Here,  M is the total mass of the solution = w + W = 13.9 g + 250 g =263.9 g

Thus the increase in the temperature (ΔT) can be calculated as:

$\Delta T = \frac{\Delta H}{M\times c}$

$\Rightarrow \Delta T = \frac{15429 J}{263.9g \times 4.184J/g.K} = 13.973$

$\Rightarrow \Delta T = T_{2} - T_{1} = 13.973$

$\Rightarrow \Delta T = T_{2} - 296K = 13.973$

$\Rightarrow T_{2} = 296K + 13.97 = 309.97K = 309.97 - 273^{\circ }C = 36.97^{\circ }C$

(∵ 0°C = 273 K)

$\Rightarrow T_{2} \approx 37^{\circ }C$

Therefore, the final temperature of the NaOH solution: T₂ = 37 °C