For the reaction 2N2O5(g) - 4NO2(g) + O2(g), if the rate of reaction …

Question

For the reaction 2N2O5(g) -> 4NO2(g) + O2(g), if the rate of reaction is r, what is the rate of formation of NO2?

Accepted Answer

D. Four times r

1. Rate $= \dfrac{1}{4}\dfrac{\Delta[NO_2]}{\Delta t} = r$.
2. Rearranging, $\dfrac{\Delta[NO_2]}{\Delta t} = 4r$.
3. So NO2 forms at four times the reaction rate.
4. Distractor B confuses the divisor with multiplier; distractor A ignores the coefficient 4.
_Source: syllabus Chemistry Ch 3 "Chemical Kinetics", p.5_

Answer

Equal to r

Answer

Half of r

Answer

Twice r

For the reaction 2N2O5(g) -> 4NO2(g) + O2(g), if the rate of reaction is r, what is the rate of formation of NO2?

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