Question

A solution of 200 mL of 1 M KOH is added to 200 mL of 1 M HCl and the mixture after attaining reaction equilibrium shows a rise in temperature by $△T_1.$ The experiment is repeated by using 50 mL of 1 M KOH and 50 mL 1 M HCl which at equilibrium shows a rise in temperature by $△T_2.$ Thus:

• A. $△T_1=△T_2$
• B. $△T_1=4\times △T_2$
• C. $△T_1=2\times △T_2$
• D. $△T_1>△T_2$

Answer 1

The correct option is A $△T_1=△T_2$

Here
Case 1
200 mL of 1 M KOH is added to 200 mL of 1 M HCl
since, volume,molarity, n factor all are same so complete neutralisation will take place and the heat released, lets say $H_1$ will be used to reaise the temperature which is $\Delta T_1$

So, we got
$H_1=mC\Delta T_1$
$H_1=400\times C\Delta T_1$ ---- eq. 1

Case 2
50 mL of 1 M KOH is added to 50 mL of 1 M HCl
since, volume ,molarity, n factor all are same so complete neutralisation will take place and the heat released, lets say $H_2$ will be used to raise the temperature which is $\Delta T_2$

So we got
$H_2=mC\Delta T_2$
$H_2=100\times C\Delta T_2$
Also, we know $H_2=\frac{H_1}{4}$ as in case 2 the reactants are one- fourth of case 1,
So , we got
$\frac{H_1}{4}=100\times C\Delta T_2$ ----- eq. 2

Comparing equation 1and 2 we got

$△T_1=△T_2$