g The combustion of ethylene proceeds by the following reaction: C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(g) If the rate of O2 is -0.23 M/s, then what is the rate (in M/s) of disappearance of C2H4?

Explanation:

C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(g)

In this reaction we see that 3 moles of O₂ reacts with one mole of C₂H₄ .

Hence rate of disappearance of O₂ is 3 times faster .

- d [O₂] / dt = - 3 d [ C₂ H₄ ] / dt

Putting the given value

.23 = 3 d [ C₂ H₄ ] / dt

d [ C₂ H₄ ] / dt  = .23 / 3

= .077 M / s

Hence the rate of disappearance of C₂ H₄ is .077 moles / s .

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