What would happen to the rate of a reaction with rate law rate = k [no]2[h2] if the concentration of no were doubled? a. the rate would be halved. b. the rate would also be doubled. c. the rate would not change. d. the rate would be four times larger.

The answer will be ( D )

The rate would be four times larger.

when the law rate = K [NO]^2[H2]

so, for example when we assume that [NO] = 2 and K = 2 and [H2] = 1 

∴ law rate = 2* 2^2 * 1 = 8

and when the [NO] is doubled that means it becomes 4

∴ law rate = 2 * 4^2 *1 = 32 

when the first law rate = 8 and after [NO] doubled = 32 

the rate would be four-time larger

Woah that’s a lot to read xd i’ll try though. i’ll get back t you if. i know the andwer