Chemistry kinetics

Question

NO reacts with H₂ in the gas phase according to the following chemical equation:

1. 2NO(g) +2H₂(g)N₂(g) +2H₂O(g)

It is observed that, when the concentration of H₂ is reduced to 1/3 of its initial value, the rate of the reaction is also reduced to 1/3 of its initial value. When the concentration of NO is multiplied by 4.92, the rate of the reaction increases by a factor of 24.2.

(a) Write the rate expression for this reaction, and give the units of the rate constant k, assuming concentration is expressed as mol L^-1 and time is in seconds.

rate =

k units =

b) If [NO] were multiplied by 4.55 and [H₂] by 2.02, what change in the rate would be observed?

The rate would increase by a factor of _____

Answer 1

A generic rate expression is Rate = k [A]^x [B]^y where [A] and [B] are the concentrations of the reactants, and x and y are called the orders of reaction. The order of a reaction must be found experimentally, and this question is implying the experiment part by giving us relative changes, saying "H₂ increased by 1/3..."

The goal is to find the exponents making the expression correct. I teach my students to use the rate expression isolated for each reactant, so

Rate = [H₂]^x

It says the concentration was increased by 1/3 and the rate increased by 1/3 - so substitute the info and solve for x. Note, the usual orders of reaction are 0, 1, and 2 - try these first :)

1/3 = [1/3]^x so x = 1 or we might say it is first order with respect to H₂

Do this again with the other condition

Rate = [NO]^y or 24.2 = [4.92]^y y = 2 or second order with respect to NO

I just teach my students to plug in 0,1,or 2 for the exponent. For a zero order, there will be no change to the rate, first order the rate change will be proportional to the concentration change, and for second order the rate change will be proportional to the square of the concentration change.

Now that we know the orders of reaction, we can substitute into the Rate equation

Rate = k [H₂]¹ [NO]²

We can find the units of the rate constant by examining the units on both sides of the equation and making them equal. The rate is mol L^-1 per second or mol L^-1 s^-1. Remember, an exponent of -1 just means it is on the bottom of the fraction. So, mol L^-1 is like saying mole per liters, mol/L, or you may know this better as Molarity!

The concentration of the [H₂] and [NO] is in Molarity, so substituting all the units into the equation would look like

mol L^-1 s^-1 = k [mol L^-1] [mol L^-1]² or combining mol L^-1 s^-1 = k mol³ L^-3

Now, I just need to figure out how k can cancel out the units on the right side to make it equal to the left side.

So k = L² mol^-2 s^-1. Again, combining mol ^3-2 L^-3+2 s^-1 or mol L^-1 s^-1 Same as the left side.

Finally, to answer part b), just substitute the values for the changes to the concentrations into the rate formula you created and calculate the change to the rate.

Rate = k [2.02] [4.55]² or 41.8 times faster.

Note: Since this problem was all relative - not giving an exact concentration or rate, just the amount it changed - I could ignore the actual value of the rate constant. If the absolute values of the concentrations and the rate were given, then I would need to plug in those values to find a value for k and use that value to solve the new rate.

Hope this helps.

Answer 2

First, we must determine the order of reaction with respect to each reactant.

For H₂, since the rate varies directly with the concentration (reduce concentration by 1/3 and also reduce rate by 1/3), this tells us the reaction is FIRST ORDER with respect to H₂.

For NO, the rate varies with the square of the concentration (4.92 squared = 24.2), this tells us the reaction is SECOND ORDER with respect to NO.

(a) Rate law: Rate = k[H₂][NO]²

For units of k, plug in the units for the knowns and solve for k:

M/s = k[M][M²]

k = M^-2s^-1

(b) rate = k[H₂][NO]² = (2.02)(4.55)² = 41.8

The rate would increase by a factor of 41.8 times