[JessMaksut]

2NO(g) + 2H₂(g) --> N₂(g) + 2H₂O(g)

The rate data for the reaction is listed below:
Experiment 1:
[NO] M = 1.2 x 10^-3
[H₂] M = 3.0 x 10^-3
Rate (M/s) = 1.08 x 10^-6

Experiment 2:
[NO] M = 2.4 x 10^-3
[H₂] M = 3.0 x 10^-3
Rate (M/s) = 4.32 x 10^-6

Experiment 3:
[NO] M = 2.4 x 10^-3
[H₂] M = 9.0 x 10^-3
Rate (M/s) = 1.30 x 10^-5

Use this data to find order of reaction with respect to NO

a) = rate2 / rate1 = k[NO]₂^x[H₂]₂^y / k [NO]₁^x[H₂]₁^y, which is equal to...
k (2.4 x 10^-3)^x(3.0 x 10^-3)^y / (1.2 x 10^-3)^x(3.0 x 10^-3)^y

Now, I understood what my professor was saying up until this point...

4.32 x 10^-6 M/s / 1.08 x 10^-6 M/s is equivilent to 2.4 x 10^-3 M^x / 1.2 x 10^-3 M ^x = 4.00 = 2^x, therefore x = 2; 2nd order with respect to NO.

When I type this into my graphing calculator, 4.32 x 10^-6 M/s / 1.08 x 10^-6 M/s DOES equal 4, but 2.4 x 10^-3 M^x / 1.2 x 10^-3 M ^x is only equal to 2. Why? Can someone show me what I'm doing wrong, whether conceptually or simply putting it into the calculator the wrong way? Thank you for your time!

Jessica

[Yggdrasil]

The rate equation for this type of reaction is:

$$ Rate = k[NO]^x [H_2]^y /$$

As you had written a good way to determine the reaction order of NO is to look at two reactions done with the same concentration of H₂ but different concentrations of NO.  The ratio of the reaction rates is give by:

$$ \frac{rate_2}{rate_1} = \frac{k [NO]_2^x[H_2]_2^y}{k [NO]_1^x [H_2]_1^y /$$

Since the concentration of H₂ is the same for both reactions 1 and 2 and since the rate constant is a constant, we can cancel those two factors and come up with a simplified expression:

$$ \frac{rate_2}{rate_1} = \left( \frac{[NO]_2}{[NO]_1} \right)^x /$$

Now, as you note, rate₂/rate₁ = 4 while [NO]₂/[NO]₁ = 2.  Since doubling the concentration of [NO] leads to a quadrupling of the rate, the reaction must be second order with respect to NO.  If doubling the concentration of NO had led to only a 2-fold increase in the rate of reaction, then the reaction would be first order with respect to NO.  Similarly, if doubling the concentration of NO had lead to an eight fold increase in the rate of reaction, the reaction would be third-order with respect to NO.