Question

1. Use the observations made from the four experiments shown in the table below to determine the rate law for the reaction 2 NO₂(g) + F₂(g)   2NO₂F(g).  Find the exponents, x and y, and the rate constant, k, and its units (you only need to use Experiment 1 values to find k).    Rate = k [NO₂]^x [F₂]^y   (6 points)

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x                                  y                                  Rate Law                                                                           k, and its units

            

Experiment	Initial Rate (mole/L.s)	Initial [NO2] (mole/L)	Initial [F2] (mole/L)
1	0.0255	0.100	0.100
2	0.0510	0.200	0.100
3	0.102	0.200	0.200
4	0.408	0.400	0.400

Answer 1

Solution:-

The rate law for the reaction is:

Rate = k [NO₂]^x [F₂]^y

Substituting the values of rate, [NO₂] and [F₂] for various sets of experiments, we have

0.0255 = k [0.100]^x [0.100]^y ..................(i) [For Experiment 1]

0.0510 = k [0.200]^x [0.100]^y ..................(ii) [For experiment 2]

0.102 = k [0.200]^x [0.200]^y ..................(iii) [For experiment 3]

0.408 = k [0.400]^x [0.400]^y ...................(iv) [For experiment 4]

Dividing eq.(ii) by eq.(i), we get

(0.0510) / (0.0255) = [0.200]^x / [0.100]^x

2 = 2^x

x = 1

Dividing eq.(iii) by eq.(ii), we get

(0.102) / (0.0510) = [0.200]^y / [0.100]^y

2 = 2^y

y = 1

Therefore, the rate law is

Rate = k [NO₂]¹ [F₂]¹

Substituting the values of rate , [NO₂ ] and [F₂] for Experiment 1 , in the rate law, we get

0.0255 mol/ L-s = k [0.100 mol/L] [0.100 mol/L]

k = (0.0255 mol/L-s) / (0.100 mol/L × 0.100 mol/L)

(Rate constant) k = 2.55 mol^-1 L s^-1

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