Problem: Consider the gas-phase reaction (constant volume) of the decomposition of nitroethane:C2H5NO2(g) → C2H4(g) + HNO2(g)At 610 K, the rate constant for this reaction is 2.1 × 10−4 s-1 initial partial pressure of C2H5NO2 is 80 torr, what will be its partial pressure after 2.5 hours?A. 12 torrB. 77 torrC. 31 torrD. 9 torrE. 18 torrF. 26 torr

FREE Expert Solution

We are asked to calculate what will be its partial pressure after 2.5 hours given the rate constant for this reaction is 2.1 × 10−4 initial partial pressure of C2H5NO2 is 80 torr.


We’re given the following first-order reaction:


C2H5NO2(g) → C2H4(g) + HNO2(g)


The integrated rate law for a first-order reaction is as follows:


ln[A]t = -kt + ln[A]0


where:

[A]t = concentration at time t 

k = rate constant

t = time, [A]0 = initial concentration


Recall that the moles of gases are proportional to the partial pressure. This means we can use the same equation and use pressure instead of concentration. 


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Problem Details

Consider the gas-phase reaction (constant volume) of the decomposition of nitroethane:

C2H5NO2(g) → C2H4(g) + HNO2(g)

At 610 K, the rate constant for this reaction is 2.1 × 10−4 s-1 initial partial pressure of C2H5NO2 is 80 torr, what will be its partial pressure after 2.5 hours?

A. 12 torr

B. 77 torr

C. 31 torr

D. 9 torr

E. 18 torr

F. 26 torr

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