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Borax

Carbon monoxi...

Question

Carbon monoxide reacts with $O_{2}$ to form ${C O}_{2} : 2 C O (g) + O_{2} (g) ⟶ 2 {C O}_{2} (g)$ .
The information regarding on this reaction is given in the table below.
[CO]mol/L
$[O_{2}]$ mol/L
Rate of reaction (mol/L min)
0.02
0.02
$4 \times 10^{- 5}$
0.04
0.02
$1.6 \times 10^{- 4}$
0.02
0.04
$8 \times 10^{- 5}$
What is the value for the rate constant for the reaction in proper related unit?

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Solution

From careful observation of the given data, it can be concluded that by doubling the concentration of $C O$ alone the rate of the reaction increases 4 times, therefore order with respect to $C O$ is two. Doubling the concentration of $O_{2}$ alone doubles the rate of the reaction, therefore order with respect to B is one.

Therefore its a third-order reaction so $R a t e = k [C O]^{2} [O_{2}]^{1}$

$4 \times 10^{- 5} = k \times (0.02)^{2} \times 0.02$

$k = 5$

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