Question

1. Write the rate law expression for the reaction \[\ce{A + B + C -> D + E}\], if the order of reaction is first, second and zero with respect to A, B and C respectively.
2. How many times the rate of reaction will increase if the concentration of A, B and C are doubled in the equation given in (a) above?

Answer 1

a. The rate law expression for the reaction:

Rate (r) = k [A]^m [B]ⁿ [C]^p

Where m is the order with respect to A (given as 1)

n is the order with respect to B (given as 2)

p is the order with respect to C (given as 0)

Thus, the rate law expression becomes:

r = k [A]¹ [B]² [C]⁰

r = k [A]¹ [B]²

b. Determine the New Rate When Concentrations are Doubled If the concentrations of A, B, and C are doubled, we can express the new concentrations as:

[A]' = 2[A] 

[B]' = 2[B]

[C]' = 2[C]

Now, substituting these new concentrations into the rate law expression:

r' = k [A]' [B]'² = k(2[A])(2[B])²

Simplify the new rate expression now, substituting the doubled concentrations:

r' = k(2[A])(2²[B]²) = k(2[A])(4[B]²)

r' = k × 8[A] [B]²

elate the New Rate to the original rate. Since the original rate r is given by:

r = k [A] [B]²

We can relate the new rate r' to the original rate:

r' = 8r