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2 A(g) + B(g) ---> 2 C(g)Rate = k [A][B]At the beginning of one trial of this reaction, [A] = 3.0 M and [B] = 1.0 M. The observed rate for the formation of C is 0.36 mol L-1 sec-1.The numerical value of k, the rate constant, is closest to0.121080.0406.0

Respuesta :

Answer:

0.12

Explanation:

The rate law of the given reaction is:-

Rate=k[A][B]

Wherem, k is the rate constant.

Given that:-

Rate = 0.36 mol/Lsec = 0.36 M/sec

[A] = 3.0 M

[B] = 1.0 M

Thus,

Applying in the equation as:-

0.36 M/sec =k × 3.0 M× 1.0 M

k = 0.12 (Ms)⁻¹

The numerical value of k, the rate constant, is closest to 0.12.

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