You are studying the reaction: NO 2 (g) + CO (g) NO (g) + CO 2 (g) at 277 K. You

Question

You are studying the reaction:

NO2 (g) + CO (g) NO (g) + CO2 (g)

at 277 K. You have collected the following data:

A plot of ln(k) vs (1/T) gave a straight line with the equation y = -2110.3 (x) + 1.0973. Additionally, the following initial rates were obtained:

(M)

(M)

1.472e-09

1.472e-09

5.888e-09

What will the rate of the reaction be when the concentration of all reactants is 2.19 M and the reaction temperature is 345 K?

Experiment [NO2]

(M)

[CO]

(M)

Rate of reaction
(M/s) 1 0.00100 0.00100

1.472e-09

2 0.00100 0.00200

1.472e-09

3 0.00200 0.00200

5.888e-09

Explanation / Answer

From the data given in the table:

a) No change in rate , even if we double the concentration of CO, so its independent of CO

zero order WRT to CO : [CO]0

b) On doubling, the Conc of NO2, the rate becomes 4 times, so its directly proptional to [NO2]2

The Rate = K [CO]0[NO2]2 over all order is 2  

1.472e-09   =K[0.001]0[0.002]2

K277 = 368000xe-09

value of   Ea at 277 K , can be calculated from equation using the slope

A plot of ln(k) vs (1/T) gave a straight line with the equation y = -2110.3 (x) + 1.0973

slope = -Ea/R = -2110.3

so Ea = 17545.034 joules at 277K

so now we need to calculate K at 345k;

ln(K2/k1) = (Ea/R){1/T1 - 1/T2}

ln(K2/368000xe-09 ) =(17545.034/8.314)(1/277- 1/345)

ln(K2/368000xe-09 )   = 1.5

ln(K2/368000xe-09 ) =ln(e1.5 )

K2/368000xe-09 =   (e1.5 )

K345 = 368000xe-09 x (e1.5 )= 368000xe-10.5

------------------------------------------

rate at 345 = K345[CO]0[NO2]2

                  = 368000xe-10.5 [2.19]o[2.19]2 = 4.7961x368000xe-10.5

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