Question

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?

Answer 1

Answer:

[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

Answer 2

Answer:

(a) The concentration of $N_{2}O_{5}$  after 5.00 min is 0.9672 mol/L

(b) The fraction of $N_{2}O_{5}$ decomposed after 5.00 min is 0.568

The problem can be solved by using first order integrated reaction.

Explanation:

(a)

Rate constant of reaction = K = $2.8\times 10^{-3}$ $s^{-1}$

Initial concentration of $N_{2}O_{5}$ = 2.24 mol/L

Assuming final concentration of $N_{2}O_{5}$ to be x mol/L

time (t) = 5.00 min  

The first order integrated equation is shown below

$\textrm{t} = \displaystyle \frac{2.303}{K}\textrm{log}\frac{\textrm{Initial concentration of } N_{2}O_{5}}{\textrm{Final concentration of } N_{2}O_{5}} \\\left ( 5.00\times 60 \right )\textrm{ sec} = \displaystyle \frac{2.303}{2.8\times 10^{-3}s^{-1}}\textrm{log}\frac{2.24 \textrm{ mol/L}}{x} \\x = 0.9672 \textrm{ mol/L}$

Concentration of $N_{2}O_{5}$ decomposed = 0.9672 mol/L

(b)

Concentration of $N_{2}O_{5}$ decomposed = $\left ( 2.24-0.9672 \right )\textrm{ mol/L} = 1.273 \textrm{ mol/L}$

Fraction of $N_{2}O_{5}$ decomposed = $\displaystyle \frac{1.273 \textrm{ mol/L}}{2.24 \textrm{ mol/L}} = 0.568$