Question

For the decomposition of hydrogen peroxide in dilute sodium hydroxide at 20°C.

2H2O2(aq)→ 2H2O(l) + O2(g)

The average rate of disappearance of H2O2 over the time period from t = 0 min to t = 477 min is found to be 4.82×10-5 M min-1.
The average rate of formation of O2 over the same time period is( _____ )M min^-1.

Answer 1

Answer: The average rate of formation of O2 over the same time period is 2.41×10-5 M min-1.

Explanation:

The equation of the reaction is given as:

2H2O2(aq)→ 2H2O(l) + O2(g)

Which implies that;

2 moles of H2O2 will produce 1 mole of O2

If H2O2 disappear at the rate of 1 mol/s, O2 would be formed at the rate of 0.5mol/s

Given the average rate of disappearance of H202 to be 4.82×10-5 M min-1.

Therefore;

O2 would be formed at the rate of;

0.5 ×4.82×10-5 M min-1. = 2.41×10-5 M min-1.

The average rate of formation of O2 over the same time period is 2.41×10-5 M min-1.

Answer 2

The average rate of formation of O₂ over the same time period is $2.41*10^{-5} M min^{-1}$.

Chemical reaction:

2 H₂O₂(aq) → 2 H₂O(l) + O₂(g)

From the reaction, we can infer that;

2 moles of H₂O₂ will produce 1 mole of O₂

If H₂O₂ disappear at the rate of 1 mol/s, O₂ would be formed at the rate of 0.5mol/s

Given:

The average rate of disappearance of H₂O₂ = $4.82*10^{-5} M min^{-1}$.

Thus,

Calculation for rate of formation:

The rate of formation of O₂ will be:

$0.5 *4.82*10^{-5} M min^{-1} = 2.41*10^{-5} M min^{-1}$

The average rate of formation of O₂ over the same time period is $2.41*10^{-5} M min^{-1}$.

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