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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 â–³T1. 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 â–³T2. Thus:

A
T1=T2
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B
T1=4×T2
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C
T1=2×T2
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D
T1>T2
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Solution

The correct option is A T1=T2
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 H1 will be used to reaise the temperature which is ΔT1

So, we got
H1=mCΔT1
H1=400×CΔT1 ---- 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 H2 will be used to raise the temperature which is ΔT2

So we got
H2=mCΔT2
H2=100×CΔT2
Also, we know H2=H14 as in case 2 the reactants are one- fourth of case 1,
So , we got
H14=100×CΔT2 ----- eq. 2

Comparing equation 1and 2 we got

T1=T2

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