Determine the rate law and the value of k for the following reaction using the data provided: 2n2o5(g) → 4 no2(g) + o2(g)0.2241.16 x 10-3 a. rate = 5.6 × 10-2 m-1s-1[n2o5]2 b. rate = 6.0 × 10-1 m-2s-1[n2o5]3 c. rate = 1.6 × 10-3 m1/2s-1[n2o5]1/2 d. rate = 1.7 × 10-2 m-1/2s-1[n2o5]3/2 e. rate = 5.2 × 10⁻3 s-1[n2o5]

2N2O5(g)> 4NO2(g) + O2(g) 
[N2O5]i (M) Initial Rate(M^-1 s^-1) 
0.093 4.84x10^-4 (1) 
0.186 9.67x10^-4 (2) 
0.279 1.45x10^-3 (3) 
From equation (1) & (2) it is evident that when [N2O5}i is doubled the initial rate is doubled, which implies the rate is directly proportional to [N2O5]. Similarly comparing equation (1) & (3) we observe that when [N2O5] is tripled the rate is also tripled. Hence the rate equation is 
Rate = k [N2O5] 
Using the data of any equation, say (1), we get 
4.84x10^-4 = k x 0.093 
OR k = 4.84x10^-4/0.093 = 5.2 x 10^-3 s-1 
Hence the rate law is 
Rate = 5.2 x 10^-3 s-1[N2O5]

The rate law and the value of k is Rate = 5.2 x 10^-3 s-1[N2O5]

 

The rate law or rate equation for a chemical reaction is an equation that links the reaction rate with concentrations or pressures of reactants and constant parameters (normally rate coefficients and partial reaction orders).

 

The correct answer between all the choices given is the last choice or letter E. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

A. volume stays constant. just took took the quiz

They carry out the exchange of gases as we breathe ( oxygen and carbon dioxide).