Question

Under some set of conditions, nitrogen gas reacts with oxygen gas to form dinitrogen monoxide. You have found that over 12.35 seconds, the concentration of dinitrogen monoxide changes from 0.23M to 3.29M. What is the rate of consumption of oxygen gas over this time period in M/sec? Report your answer to 4 decimal places.

Answer 1

The balanced chemical reaction for the reaction for the formation of N2O is

2N2(g) + O2(g) ------- >2 N2O(g)

The rate for the above reaction can be written. as

rate = - 1/2 x {d[N2(g)] / dt} = - 1x{d[O2(g)] / dt} = +1/2 x {d[N2O(g)] / dt}

=>   - 1x{d[O2(g)] / dt} = +1/2 x {d[N2O(g)] / dt}

=>  - 1x{d[O2(g)] / dt} = +1/2 x {(3.29M - 0.23M) / 12.35s} = 0.1239 M.s^-1

=> {d[O2(g)] / dt} = - 0.1239 M.s^-1

Hence rate of consumption of oxygen gas is 0.1239 M.s^-1 (answer)