The rate constant for the second order reaction 2 NO2 - 2 NO + O2 is 0.24 M's. If a 1.38 M sample of NO2 reacts for 47.7 seconds, what concentration of NO2 wil remain? Report your answer to two decimal places. |Al = - kt +|Alo InJAl = - kt + InlAlo TAI = kt + TAlo

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The rate constant for the second order reaction 2 NO2 - 2 NO + O2 is 0.24 M's. If a 1.38 M sample of NO2 reacts for 47.7 seconds, what concentration of NO2 wil
remain?
Report your answer to two decimal places.
|Al = - kt +|Alo
InJAl = - kt + InlAlo
TAI
= kt +
TAlo
Transcribed Image Text:The rate constant for the second order reaction 2 NO2 - 2 NO + O2 is 0.24 M's. If a 1.38 M sample of NO2 reacts for 47.7 seconds, what concentration of NO2 wil remain? Report your answer to two decimal places. |Al = - kt +|Alo InJAl = - kt + InlAlo TAI = kt + TAlo
Expert Solution
Step 1

For a second order reaction, rate equation is given as : A0 = A + kt

Initial concentration (A0) = 1.38 M,

Rate constant (k) = 0.24 M-1 s-1

Time (t) = 47.7 s

Substiting the values in the integrated rate equation for 2nd order is we get,

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