The decomposition reaction of n2o5 in carbon tetrachloride is 2n2o5−→−4no2+o2. the rate law is first order in n2o5. at 64 °c the rate constant is 4.82×10−3s−1. (a) write the rate law for the reaction. (b) what is the rate of reaction when [n2o5]=0.0240m? (c) what happens to the rate when the concentration of n2o5 is doubled to 0.0480 m? (d) what happens to the rate when the concentration of n2o5 is halved to 0.0120 m?

(a) rate =  4.82 x 10⁻³s⁻¹  [N2O5]

(b) rate =   1.16 x 10⁻⁴  M/s

(c) rate =   2.32 x 10⁻⁴ M/s

(d) rate =   5.80 x 10⁻⁵ M/s

Explanation:  

We are told the rate law is first order in N₂O₅, and its rate constant is 4.82 x 10⁻³s⁻¹ . This means the rate is proportional to the molar concentration   of   N₂O₅, so

(a) rate = k [N2O5] = 4.82 x 10⁻³s⁻¹ x [N2O5]

(b) rate = 4.82×10⁻³s⁻¹  x 0.0240 M =  1.16 x 10⁻⁴ M/s

(c) Since the reaction is first order if the concentration of  N₂O₅ is double the rate will double too:  2 x   1.16 x 10⁻⁴ M/s = 2.32 x 10⁻⁴ M/s

(d) Again since the reaction is halved to 0.0120 M, the rate will be halved to

1.16 x 10⁻⁴ M/s / 2 =  5.80 x 10⁻⁵ M/s

c. 4.54 x 10–12 j happens to be correct