163k views
1 vote
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

1 Answer

3 votes

Answer:

0.12

Step-by-step 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.

User Radu Andrei
by
7.5k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.