Question

A reaction with stoichiometric equation 1/2A + B = R + 1/2S has the following rate expression –rA = 2CA 0.5 CB. What is the rate expression for this reaction if the stoichiometric equation is written as A + 2B = 2R + S?

Answer #1

Reaction equation: ½ A + B ====> R + ½ S

Rate of disappearance of A, -r_{A} =
k[A]^{0.5}[B]

Can also be written as -r_{A} = k
C_{A}^{0.5} C_{B}

rate expression given for the above reaction equation =
-r_{A} = 2 C_{A}^{0.5} C_{B}.

So the rate constant, k, of the reaction = 2

If the stoichiometric equation is written as A + 2B = 2R + S

Then the rate of disappearance of A will be, -r_{A} =
k[A][B]^{2}

Can also be written as -r_{A} = k
C_{A} C_{B}^{2}

Since rate constant, k, of the reaction = 2

The rate of disappearance of A will be, -r_{A} =
2[A][B]^{2}

Suggest a mechanism for below reaction: 2A+B → R when at the
beginning the reaction rate is linearly proportional to
concentration of CA, but at the end is proportional to CA2 .

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Experiement #
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1
0.300
0.150
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