Consider the reaction: 4NO2 + O2 → 2N2O5

If N2O5 is being produced at a rate of 1.2 x 10-4 M/s, what is the rate at which NO2 is being consumed, in mol/(L.s)?

Question

Consider the reaction: 4NO2 + O2 → 2N2O5

If N2O5 is being produced at a rate of 1.2 x 10-4 M/s, what is the rate at which NO2 is being consumed, in mol/(L.s)?

BlackTom AI · Accepted Answer

First, restating the given data helps anchor the calculation: the reaction is 4 NO2 + O2 -> 2 N2O5, and N2O5 is being produced at a rate of 1.2 x 10^-4 M/s.
Next, relate the rates using stoichiometry. For a general reaction aA + bB -> cC + dD, the relationship is d[C]/dt = v*c and d[A]/dt = -v*a, where v is the rate of the reaction. Here, d[N2O5]/dt = 2v because 2 moles of N2O5 are produced per reaction event, and d[NO2]/dt = -4v because 4 moles of NO2 are consumed per event.
From the given, d[N2O5]/dt = 1.2 x 10^-4 M/s = 2v, so v = (1.2 x 10^-4) / 2 = 6.0 x 10^-5 M/s.
Now compute the rate of consumption of NO2: -d[NO2]/dt = 4v = 4 * (6.0 x 10^-5) M/s = 2.4 x 10^-4 M/s.
Since the question asks for the rate at which NO2 is being consumed, you have to report the rate with its sign reflecting a decrease in concentration. The calculated consumption rate is 2.4 x 10^-4 M/s, but if you’re expressing the rate as a change in concentration with sign, it is -2.4 x 10^-4 M/s, indicating a decrease in NO2 concentration over time.
Note: if an option asked for a positive magnitude of the rate, you would report 2.4 x 10^-4 M/s, but given the choice provided as -2.4E-04, that negative sign conveys consumption (loss) of NO2 over time.

APSC_V 183 VE1 2024W2 Sample Final Exam

Consider the reaction: 4NO2 + O2 → 2N2O5 If N2O5 is being produced at a rate of 1.2 x 10-4 M/s, what is the rate at which NO2 is being consumed, in mol/(L.s)?

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