Consider the following reaction:

a + b + c right arrow d

the rate law for this reaction is as follows:

rate = k time fraction [a][c]^2 over [b]^1/2

suppose the rate of the reaction at certain initial concentrations of a, b, and c is 1.12×10-2 m/s.

what is the rate of the reaction if the concentrations of a and c are doubled and the concentration of b is tripled?

rate 2 = ? m/s

The rate of the reaction when concentrations are changed is

Explanation:

For the given chemical reaction:

The given rate law of the reaction follows:

We are given:

When the concentrations are changed:

New concentration of A = 2A  (Concentration is doubled)

New concentration of C = 2C   (Concentration is doubled)

New concentration of B = 3B     (Concentration is tripled)

The new rate law expression becomes:

Hence, the rate of the reaction when concentrations are changed is

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.

dynamic equilibrium is the answer.