What would happen to the rate of a reaction with rate law rate = k no2 h2 if the concentration of h2 were halved?

The question is not formatted correctly. The proper question should be

What would happen to the rate of a reaction with rate law

rate=k [NO]2[H2]

if the concentration of H2 were halved?

Rate would be halved.

Explanation:-

Let originally [NO]= 1 and [H2]=1.

Then original rate= k [1]2 [1] = k.

Now if [H2] is halved then new [H2] = 1/2

While [NO] remains same.

Then

new rate = k [1]2 [1/2] = k/2

Thus the rate is halved.

43.452

step-by-step explanation:

284 times .153 = 43.452