When equal volume of 1M HCl and 1M H$_2$SO$_4$ are separately neutralized by excess volume of 1M NaOH solution, X and Y kJ of heat is liberated respectively. The value of Y/X is ____.
Correct Answer: 2
Approach Solution - 1
To solve the problem, we must first understand the concept of neutralization enthalpy. When a strong acid is neutralized by a strong base, the reaction releases heat. The standard enthalpy change for neutralization of a strong acid by a strong base is approximately 57 kJ/mol.
Consider the reactions:
1. For HCl: HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)
2. For H2SO4: H2SO4(aq) + 2NaOH(aq) -> Na2SO4(aq) + 2H2O(l)
In the first reaction, 1 mole of HCl neutralizes 1 mole of NaOH. Since 1M HCl is used, and mixed in equal volume with NaOH, 1 mole of NaOH will react, releasing 57 kJ of heat per mole. Thus, X = 57 kJ.
In the second reaction, 1 mole of H2SO4 neutralizes 2 moles of NaOH. Hence, for 1 mole of H2SO4 mixed with NaOH, 2 moles of NaOH react, liberating 57 kJ × 2 = 114 kJ of heat. Therefore, Y = 114 kJ.
The ratio of heat liberated Y/X = 114/57 = 2.
This value, 2, matches the expected range of (2,2), confirming our solution's consistency and correctness.
Approach Solution -2
Neutralization reactions:
\( \mathrm{H^+ + OH^- \rightarrow H_2O} \quad (\text{from HCl: } X), \)
\( \mathrm{2H^+ + 2OH^- \rightarrow 2H_2O} \quad (\text{from } H_2SO_4: Y). \)
From the reactions:
\( Y = 2X \implies \frac{Y}{X} = 2. \)
Final Answer: 2.
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