Rate laws: reaction rate as a function of reactant concentrations T02/S02 At a c

Question

Rate laws: reaction rate as a function of reactant concentrations T02/S02 At a certain temperature the following data were collected for the reaction: CO(g) + Cl2(g) COCl(g) + Cl(g) [co](M) [Cl2](M) Initial rate (M/s) 1.00-102 1.00-102 2.00 102 1.00 102 1.32 104 3.00 102 1.00 102 1.98-104 1.00 102 2.00 102 2.64*104 .00102 3.00 102 1.19-10s 6.60*10 What is the rate law for this reaction? O rate = k[CO] [Cl2]2 rate k[Co2[C212 rate = k[CO]2[OJ2 O rate = k[CO]2[Cl2] O rate = k[CO] [Cl2]4 What is the value of the rate constant for the reaction? 2.mol-2.min-1

Explanation / Answer

In the table, we can see that as the concentration of CO is doubled, keeping the concentration of Cl2 same, the rate doubled and similarly by increasing the concentration by triple times, the rate becomes three times. So, rate is proportional to [CO]. But in case of Cl2, when the concentration of Cl2 is doubled keeping the concentration of CO same, the rate becomes 4 times. So, the rate is proportional to [Cl2]2.

So, rate = k [CO][Cl2]2

1st option is correct.

To calculate rate constant:

rate = k [CO][Cl2]2

rate = 6.60 * 103 M/s

[CO] = 1.00 * 102 M

[Cl2] = 1.00 * 102 M

So, rate = 6.60 * 103 = k (1.00 * 102 M) (1.00 * 102 M)2

k = (6.60 * 103)/[(1.00 * 102 M) (1.00 * 102 M)2 ]

k = 6.60 * 10-3 M-2 s-1 = 6.60 * 10-3 L2mol-2 s-1

rate constant = 6.60 * 10-3 L2 mol-2 s-1

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