Rate Law

asked by @jeremyb10 • about 1 year ago • Chemistry • 5 pts

I am given Rate = k[N2]^2[H2]^2

I am told at a certain concentration of N2 and H2, the INITIAL rate of the reaction is 40.0 M/s. What would the initial rate of the reaction be if the concentration of N2 were halved?

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1 answer

It would be 10.0 M/s.

answered by @nikaa18 • about 1 year ago
  • Hi there! I'd just like to elaborate since it is a second order for N2 you the effects of changing the concentration will be squared. Meaning if you make it half the original the rate would be affected by the square of that which is 1/4 which makes sense why the answer should be 10.0 M/s. I hope this helps! Leo commented about 1 year ago