Question:

If 100 mole of \(\mathrm{H_2O_2}\) decompose at 1 bar and 300 K, the work done (kJ) by one mole of \(\mathrm{O_2(g)}\) as it expands against 1 bar pressure is :
\(2\mathrm{H_2O_2(l)} \rightleftharpoons 2\mathrm{H_2O(l)} + \mathrm{O_2(g)}\)
(\(R = 8.3\ \mathrm{J\,K^{-1}\,mol^{-1}}\))

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Only \(\mathrm{O_2}\) is gaseous, so work \(= \Delta n_g RT\) where \(\Delta n_g\) is the moles of \(\mathrm{O_2}\) formed from 100 mol of \(\mathrm{H_2O_2}\).
Updated On: Nov 11, 2026
  • 62.25
  • 124.50
  • 249.00
  • 498.00
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The Correct Option is B

Solution and Explanation

The reaction gives 1 mol \(\mathrm{O_2}\) per 2 mol \(\mathrm{H_2O_2}\), so 100 mol \(\mathrm{H_2O_2}\) give 50 mol \(\mathrm{O_2}\); liquids contribute negligible volume.
Work done by the gas \(= p\Delta V = \Delta n_g RT = 50\times8.3\times300 = 124500\) J \(= 124.5\) kJ.
Answer: (2) 124.50
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