The first-order rate constant for the decomposition of N2O5, 2N2O5(g) yields 4NO2(g) at 70∘C is 6.82 x 10-3 s-1 . Suppose we start with 2.30×10−2 mol of N2O5(g) in a volume of 1.8 L . How many moles of N2O5 will remain after 7.0 minutes?
The first-order rate constant for the decomposition of N2O5, 2N2O5(g) yields 4NO2(g) at 70∘C is 6.82 x 10-3 s-1 . Suppose we start with 2.30×10−2 mol of N2O5(g) in a volume of 1.8 L . How many moles of N2O5 will remain after 7.0 minutes?
Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter13: Chemical Kinetics
Section: Chapter Questions
Problem 13.57QE: The decomposition of ozone is a second-order reaction with a rate constant of 30.6 atm1 s1 at 95 C....
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The first-order rate constant for the decomposition of N2O5,
2N2O5(g) yields 4NO2(g)
at 70∘C is 6.82 x 10-3 s-1 . Suppose we start with 2.30×10−2 mol of N2O5(g) in a volume of 1.8 L .
How many moles of N2O5 will remain after 7.0 minutes?
Expert Solution
Step 1
Given that :
Rate constant = 6.82 x 10-3 s-1
Initial concentration = 2.30×10−2
Volume = 1.8 L
We have to calculate the moles of N2O5 will remain after 7.0 minutes.
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