Question

200 mL of 0.2 M HCl is mixed with 300 mL of 0.1 M NaOH. The molar heat of neutralization of this reaction is \(-57.1\) kJ. The increase in temperature in \(^\circ C\) of the system on mixing is \(x \times 10^{-2}\). The value of \(x\) is _________. (Nearest integer)
[ Given : Specific heat of water \(= 4.18 \, \text{J g}^{-1} \text{K}^{-1}\)
Density of water \(= 1.00 \, \text{g cm}^{-3}\) ]
(Assume no volume change on mixing)

Hint:

Always identify the limiting reagent in neutralization problems. The heat released is proportional to the number of moles of water formed (or the moles of limiting \(H^+\) or \(OH^-\)).

Answer 1

Correct Answer: 82

To solve this problem, we will calculate the heat of neutralization and then determine the temperature increase in the system.

Step 1: Calculate Moles of HCl and NaOH
We start by determining the moles of reactants:

• Moles of HCl = 0.2 M × 0.2 L = 0.04 mol
• Moles of NaOH = 0.1 M × 0.3 L = 0.03 mol

Since NaOH is the limiting reagent, 0.03 mol of NaOH will react completely.

Step 2: Heat of Neutralization
The reaction between HCl and NaOH releases heat according to the equation:
HCl + NaOH → NaCl + H₂O + Heat
The molar heat of neutralization is \(-57.1\) kJ/mol. Thus, the total heat (q) released is:

q = 0.03 mol × \(-57.1\) kJ/mol = -1.713 kJ
Convert to Joules: q = -1713 J

Step 3: Calculate Temperature Increase
We know:

• Density of water = 1 g/cm³
• Total volume = 200 mL + 300 mL = 500 mL = 500 g (assuming density 1 g/mL)
• Specific heat (c) of water = 4.18 J/g°C

Using q = mcΔT, we solve for ΔT:

ΔT = q / (mc) = -1713 J / (500 g × 4.18 J/g°C) = -0.819 °C

Express the temperature increase as \(x \times 10^{-2}\):
\(x \times 10^{-2} = 81.9 \Rightarrow x = 82\)

Therefore, the value of \(x\) is 82, which is within the specified range (82, 82).