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?
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.
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