Question

When 600 mL of 0.2 M HNO₃ is mixed with 400 mL of 0.1 M NaOH solution in a flask, the rise in temperature of the flask is ________ × 10^–2 °C.
(Enthalpy of neutralisation = 57 kJ mol^–1 and Specific heat of water = 4.2 JK^–1 g^–1)
(Neglect heat capacity of flask)

Answer 1

Correct Answer: 54

To find the rise in temperature, we first calculate the moles of HNO₃ and NaOH. Using the formula Moles = Molarity × Volume in L:

Moles of HNO₃ = 0.2 M × 0.6 L = 0.12 mol

Moles of NaOH = 0.1 M × 0.4 L = 0.04 mol 

Since NaOH is the limiting reactant, only 0.04 mol of HNO₃ reacts. The heat change (q) from this neutralization is q = −ΔH × moles reacted = −57 kJ/mol × 0.04 mol = −2.28 kJ

This heat is absorbed by the water produced. Mass of water = Total volume = 600 mL + 400 mL = 1000 g (assuming the density of water = 1 g/mL).

Using q = mcΔT, where m = 1000 g, c = 4.2 Jg^–1K^–1, and ΔT is the temperature change:

2.28 kJ = 2280 J = 1000 g × 4.2 Jg^–1K^–1 × ΔT

Solve for ΔT: ΔT = 2280 J / 4200 J/K = 0.5429 K

Expressing the temperature change in the desired form: 0.5429 K = 54.29 × 10^–2°C

Thus, the rise in temperature is 54 × 10^–2°C, fitting the expected range (54,54).

Answer 2

\(\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}\)
m moles 120 40 – 80 – 40

Heat liberated from reaction
= 40 × 10^–3× 57 × 10³ J …(1)

Heat gained by solution = \(mC\Delta T\)
Mass of solution \(m = V × d = 1000 × 1 = 1000 g\)
Heat gained by solution \(= 1000 × 4.2 × ΔT …(2)\)
From eq (1) and eq (2)
Heat liberated = Heat gained
\(40 × 10–3 × 57 × 103 = 1000 × 4.2 × ΔT\)
\(ΔT = 54 ×\) 10^–2 \(°C\)
\(ΔT = 54°C\) (Rounded off to the nearest integer)
So, the answer is 54°C.