Question

What would happen to the rate of a reaction with rate law rate = k [NO]^2[H2] if

the concentration of NO were halved?
A. The rate would be four times larger.

B. The rate would also be halved.

C. The rate would be one-fourth.

D. The rate would be doubled.

Answer 1

C. The rate would be one-fourth.

Another option would be:

The rate would be four times larger

Depending on what class you're taking.

Step-by-step explanation:

Answer 2

The rate of a reaction would be one-fourth.

Further explanation

Given

Rate law-r₁ = k [NO]²[H2]

Required

The rate of a reaction

Solution

The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.

Can be formulated:

Reaction: aA ---> bB

`\large{\boxed{\boxed{\bold{v~=~-(\Delta A)/(\Delta t)}}}`

or

`\large{\boxed{\boxed{\bold{v~=~+(\Delta B)/(\Delta t)}}}`

The concentration of NO were halved, so the rate :

`\tt r_2=k[(1)/(2)No]^2[H_2]\\\\r_2=(1)/(4)k.[No]^2[H_2]\\\\r_2=(1)/(4)r_1`