The first-order rate constant for the decomposition of n2o5, 2n2o5 (g) →4no2 (g) + o2 (g) at 70∘c is 6.8210-3 s-1. suppose we start with 2.0010-2 mol of n2o5 (g) in a volume of 2.3 l. you may want to reference (page) section 14.4 while completing this problem. part a how many moles of n2o5 will remain after 4.0 min?,

Question

Chemistry

Jarrett

14 November, 08:53

The first-order rate constant for the decomposition of n2o5, 2n2o5 (g) →4no2 (g) + o2 (g) at 70∘c is 6.8210-3 s-1. suppose we start with 2.0010-2 mol of n2o5 (g) in a volume of 2.3 l. you may want to reference (page) section 14.4 while completing this problem. part a how many moles of n2o5 will remain after 4.0 min?,

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Roberto Mcconnell · Answer 1

The rate constant for 1st order reaction is

K = (2.303 / t) log (A0 / A)

Where, k is rate constant

t is time in sec

A0 is initial concentration

(6.82 * 10-3) * 240 = log (0.02 / A)

1.63 = log (0.02 / A)

-1.69 - log A = 1.63

Log A = - 0.069

A = 0.82

Hence, 0.82 mol of A remain after 4 minutes.