Question

The reaction of nitric oxide, NO, with hydrogen, H₂, produces nitrogen, N₂, and water, H₂O, according to the equation

2 NO + 2 H₂ ? N₂ + 2 H₂O

A student obtained the following data:

Experiment	Initial [H2]/ mol L?	Initial [NO]/ mol L?	Initial rate/ mol L?1 s?1
1	0.20	0.20	3.2 x 10-3
2	0.40	0.20	1.3 x 10-2
3	0.20	0.40	6.4 x 10-3

(a) Determine the most probable rate equation for the reaction, with v for the reaction rate and k for the rate coefficient.

(b) Evaluate the numerical value (with units) of the rate coefficient.

(c) Choose the appropriate matches below. Note that some of the units have to be expressed in a particular way in the drop down menus, e.g. L⁴ shows as L4, s^-1 looks like s-1, L⁴ s^-1 looks like L4 s-1.

Background reading:

the Chemguide concentration dependence page (38kB)

the Chem guide orders and mechanism page (28kB)

Energy section Powerpoint lecture notes

Order with respect to [NO]	Answer 1Choose...0.4 L2 mol-2 s-12.0 L2 mol-2 s-121reaction rate0.4 s-10.08 L2 mol-2 s-1
Order with respect to [H2]	Answer 2Choose...0.4 L2 mol-2 s-12.0 L2 mol-2 s-121reaction rate0.4 s-10.08 L2 mol-2 s-1
v	Answer 3Choose...0.4 L2 mol-2 s-12.0 L2 mol-2 s-121reaction rate0.4 s-10.08 L2 mol-2 s-1
k	Answer 4Choose...0.4 L2 mol-2 s-12.0 L2 mol-2 s-121reaction rate0.4 s-10.08 L2 mol-2 s-1

Answer 1

2 NO + 2 H2 ? N2 + 2 H2O

For the above equation

Let rate law

r = k [NO]^a [H2]^b

From trial 1 and 2

r1/r2 = ( [NO]1^a [H2]1^b) / ( [NO]2^a [H2]2^b)

(3.2 *10^-3 / (1.3*10^-2) = (0.20/0.40)^b

0.246 = 0.5^b

Take natural logarithm on both sides

ln 0.246 = b ln 0.5

b = 2

Similarly from trial 1 and 3

r1/r3 = ( [NO]1^a [H2]1^b) / ( [NO]3^a [H2]3^b)

(3.2 *10^-3 / (6.4*10^-3) = (0.20/0.40)^a

0.5 = 0.5^a

a = 1

Rate law

r = k [NO] [H2]²

Part b

From trial 1

r1 = k [NO]1 [H2]1²

3.2 x 10^-3 mol L?1 s?1 = k x 0.20 M x 0.20 x 0.20 M2

3.2 x 10^-3 M s?1 = k x 0.008 M3

k = (3.2 x 10^-3 M s?1) / (0.008 M3)

k = 0.40 M-2 s-1

k = 0.40 L2 mol-2 s-1

Part C

Order with respect to [NO] = 1

Order with respect to [H2] = 2

Rate law = r = k [NO] [H2]²

From trial 1

r = 0.40 x 0.20 x 0.20 x 0.20 = 0.0032 mol L?1 s?1

For trial 2

r = 0.40 x 0.20 x 0.40 x 0.40 = 0.0128 mol L?1 s?1

For trial 3

r = 0.40 x 0.40 x 0.20 x 0.20 = 0.0064 mol L?1 s?1

Rate constant k = 0.40 L2 mol-2 s-1