For the decomposition of gaseous dinitrogen pentaoxide, 2 N2O5(g)→4 NO2(g) + O2(g) the rate constant is k = 2.8 × 10−3 s−1 at 60°C. The initial concentration of N2O5 is 2.24 mol/L. (a) What is [N2O5] after 5.00 min? mol L (b) What fraction of the N2O5 has decomposed after 5.00 min?

[N2O5] = 0.967 mol/L

Fraction which has decomposed = 0.568

Explanation:

Step 1: Data given

k = 2.8 * 10^−3/s at 60°C

The initial concentration of N2O5 is 2.24 mol/L

Step 2: Decomposition of N2O5

What kind of order ?

The fact that the units of k are reciprocal seconds means it is first order.

Rate = k[A]

Integrated rate law:

[A] = [Ao] e^(-kt)  

[A] = 2.24 mol/L * e^(-2.8*10^-3/s * 300s)

[A] = 0.967 mol/L

Step 3: Calculate the fraction of N2O5 that has decomposed after 5.00 min

Fraction which has decomposed = (1 - 0.967 / 2.24) = 0.568

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43.452

step-by-step explanation:

284 times .153 = 43.452