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
Solution and Explanation
The problem requires understanding neutralization enthalpy. Strong acid-strong base reactions generate heat. The standard enthalpy change for this process is approximately 57 kJ/mol.
Consider the following reactions:
1. HCl neutralization: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
2. H2SO4 neutralization: H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
In reaction 1, 1 mole of HCl reacts with 1 mole of NaOH. If 1M HCl is mixed with an equal volume of NaOH, 1 mole of NaOH will react, releasing 57 kJ. Therefore, X = 57 kJ.
In reaction 2, 1 mole of H2SO4 reacts with 2 moles of NaOH. Consequently, when 1 mole of H2SO4 is mixed with NaOH, 2 moles of NaOH will react, releasing 57 kJ × 2 = 114 kJ. Thus, Y = 114 kJ.
The ratio of heat liberated is Y/X = 114/57 = 2.
This calculated ratio of 2 falls within the expected range of (2,2), validating the accuracy of the solution.
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