Question

Determine the rate law and the value of k for the following reaction using the data provided.

2 N2O5(g) → 4 NO2(g) + O2(g)

[ N2O5]i (M)

Initial Rate (M-1s-1)

​

0.033​

1.84 × 10-4

​

0.066

7.37 × 10-4

​

0.099

1.66× 10-3

• A.

  Rate = 5.6 × 10-1 M-1/2s-1[N2O5]3/2

• B.

  Rate = 1.7 × 10-1 M-1s-1[N2O5]2

• C.

  Rate = 1.7 × 10-1M1/2s-1[N2O5]1/2

• D.

  Rate = 6.0 × 10-1 M-2s-1[N2O5]3

• E.

  Rate = 5.2 × 10^⁻1 s-1[N2O5]

Answer 1

Say the rate equation is,

rate = k. N205)

where, m is the order of the reaction, k is the rate constant, and [ ] stands for concentration.

from the three data,

$1.84\times 10^{-4}\, M^{-1}.s^{-1}=k.(0.033\, M)^m..........(i)$

$7.37\times 10^{-4}\, M^{-1}.s^{-1}=k.(0.066\, M)^m..........(ii)$

$1.66\times 10^{-3}\, M^{-1}.s^{-1}=k.(0.099\, M)^m..........(iii)$

(ii) / (i) gives,

$\frac{7.37\times 10^{-4}}{1.84\times 10^{-4}}=\frac{k.0.066^m}{k.0.033^m}$

$\Rightarrow 4=2^m$

$\Rightarrow 2^2=2^{m}$

$\therefore m=2$

Thus, rate = k [N2O5]2

Now, applying this in equation (iii) we get,

$1.66\times 10^{-3}\, M^{-1}.s^{-1}=k.(0.099\, M)^{2}$

$k=\frac{1.66\times 10^{-3}\, M^{-1}.s^{-1}}{0.099^2\, M^2}=1.7\times 10^{-1}M^{-3}.s^{-1}$

Thus,

$rate=1.7\times 10^{-1}\, M^{-3}.s^{-1}\: [N_2O_5\, M]^2=1.7\times 10^{-1}\, M^{-1}.s^{-1}\: [N_2O_5]^2$

Thus, asnwer is B.