Question

When 400 mL of 0.2 M H$_2$SO$_4$ solution is mixed with 600 mL of 0.1 M NaOH solution, the increase in temperature of the final solution is ________ $\times 10^{-2}$ K. (Round off to the Nearest Integer).
[Use : H$^+$(aq) + OH$^-$(aq) $\rightarrow$ H$_2$O(l) ; $\Delta_r H$ = -57.1 kJ mol$^{-1}$
Specific heat of H$_2$O = 4.18 J K$^{-1}$ g$^{-1}$
density of H$_2$O = 1.0 g cm$^{-3}$
Assume no change in volume of solution on mixing.]

Hint:

In neutralization calorimetry, always find the limiting reactant (either H$^+$ or OH$^-$) first, as the amount of heat released depends on the number of moles of water formed.

Answer 1

Correct Answer: 82

This is a calorimetry problem involving a neutralization reaction.
Step 1: Calculate the moles of H$^+$ and OH$^-$.
Moles of H$_2$SO$_4$ = Molarity $\times$ Volume = 0.2 mol/L $\times$ 0.400 L = 0.08 mol.
Since H$_2$SO$_4$ is a strong acid, it provides 2 H$^+$ ions per molecule.
Moles of H$^+$ = 2 $\times$ 0.08 mol = 0.16 mol.
Moles of NaOH = Molarity $\times$ Volume = 0.1 mol/L $\times$ 0.600 L = 0.06 mol.
Since NaOH is a strong base, it provides 1 OH$^-$ ion per molecule.
Moles of OH$^-$ = 0.06 mol.
Step 2: Determine the limiting reactant and calculate the heat released.
The neutralization reaction is H$^+$ + OH$^-$ $\rightarrow$ H$_2$O. The stoichiometry is 1:1.
Since moles of OH$^-$ (0.06)<moles of H$^+$ (0.16), OH$^-$ is the limiting reactant.
Moles of water formed = 0.06 mol.
Heat released, q = (moles of limiting reactant) $\times$ (enthalpy of neutralization)
q = 0.06 mol $\times$ 57.1 kJ/mol = 3.426 kJ = 3426 J.
Step 3: Calculate the temperature change.
Total volume of the final solution = 400 mL + 600 mL = 1000 mL.
Assuming the density of the solution is that of water (1.0 g/mL), the total mass of the solution is m = 1000 g.
The specific heat of the solution is given as s = 4.18 J K$^{-1}$ g$^{-1}$.
Using the formula q = ms$\Delta$T:
3426 J = (1000 g) $\times$ (4.18 J K$^{-1}$ g$^{-1}$) $\times$ $\Delta$T.
$\Delta$T = $\frac{3426}{1000 \times 4.18} = \frac{3426}{4180} \approx 0.8196$ K.
Step 4: Express the answer in the required format.
The question asks for the answer in the form ______ $\times 10^{-2}$ K.
$\Delta$T = 0.8196 K = 81.96 $\times 10^{-2}$ K.
Rounding to the nearest integer, the value is 82.