The reaction 2NO (g) + 2H2 -> N2 (g) + 2H2O (g) was found to follow the rate law, rate = k[NO]^2[H2]. By what factor will the rate of reaction increase when the pressure of NO gas is increased from 2.0 atm to 3.0 atm? Assume all other conditions are held constant.

I'm having a really hard time with rate laws. The only thing I know about this problem is that increasing the pressure will increase the rate of reaction, but I don't know how to get the answer. The answer is supposed to be 2.25.

You're making it much too hard.

How mcuh is rate increased when NO is doubled? That's
rate = k[NO]^2[H2]. When NO is doubled rate is 4x; i.e., 2^2 = 4.
When tripled rate is 9x; i.e., 3^2 = 9.
What's the ratio of 9/4?

To determine the effect of changing the pressure of NO gas on the rate of reaction, we need to examine a relationship between the concentrations of the reactants and the rate of reaction.

Given the rate law you provided:
rate = k[NO]^2[H2]

We can assume that the concentration of H2 is held constant.

Let's analyze the effect of increasing the pressure of NO gas from 2.0 atm to 3.0 atm:

According to the ideal gas law, the pressure of a gas is directly proportional to its concentration at constant temperature:

P = nRT/V

Assuming the volume (V), temperature (T), and number of moles (n) of NO gas remain constant, we can write:

P1/V = [NO]1
P2/V = [NO]2

As the volume is constant, we can simplify:

[NO]1 = P1
[NO]2 = P2

Now let's calculate the ratio of rates using the given concentrations:

(rate2 / rate1) = ([NO]2)^2 / ([NO]1)^2

Plugging in the values, we get:

(rate2 / rate1) = (3.0^2) / (2.0^2)
(rate2 / rate1) = 9 / 4

Simplifying the equation, we find:

(rate2 / rate1) = 2.25

Therefore, when the pressure of NO gas is increased from 2.0 atm to 3.0 atm, the rate of the reaction will increase by a factor of 2.25.

To determine the factor by which the rate of reaction will increase when the pressure of NO gas is increased, we need to use the given rate law and the concept of reaction orders.

1. Begin by writing the balanced equation for the reaction:
2NO (g) + 2H2 -> N2 (g) + 2H2O (g)

2. Identify the reaction orders for each reactant. In this case, the reaction order for NO is given as 2, while the reaction order for H2 is also given as 1 (as it is not explicitly mentioned).

3. The rate law equation for this reaction is rate = k[NO]^2[H2].

4. To find the factor by which the rate increases when the pressure of NO gas is changed, we can use the relationship between concentration and pressure. This is known as the ideal gas law: PV = nRT.

5. Assuming constant temperature and volume, we can approximate P as being directly proportional to the concentration of NO gas in moles per liter (M). Therefore, if the pressure doubles, the concentration also doubles.

6. Let's consider the initial pressure of NO as P1 = 2.0 atm and the final pressure as P2 = 3.0 atm.

7. Since the pressure is directly related to the concentration, we can say that [NO]1/[NO]2 = P1/P2.

Substituting the given values, [NO]1/[NO]2 = 2.0 atm / 3.0 atm = 2/3.

8. Now, we can use the given rate law and the relationship between concentration and rate:

rate1 / rate2 = ([NO]1 / [H2]1)^2 / ([NO]2 / [H2]2)^2

Since [H2] is held constant, we can remove it from the equation:

rate1 / rate2 = ([NO]1)^2 / ([NO]2)^2

rate1 / rate2 = (2/3)^2 / 1^2 = (4/9) / 1 = 4/9

rate2 / rate1 = 9/4

Therefore, the rate of reaction increases by a factor of 9/4 or 2.25 when the pressure of NO gas is increased from 2.0 atm to 3.0 atm.

Hence, the answer is indeed 2.25.