Q1(a) [10 Marks] The table below shows the data obtained...

Q1(a) [10 Marks] The table below shows the data obtained from a kinetics experiment for the reaction 2NO (g) + Cl? (g) ? 2 NOCl (g) by determining the initial rate of loss of NO at different reactant concentrations at 25pC: Experiment | [NO] / M | [Cl?] / M | Rate of loss of NO / Ms?¹ 1 | 0.100 | 0.100 | 3.00 x 10?³ 2 | 0.100 | 0.150 | 4.50 x 10?³ 3 | 0.150 | 0.100 | 6.75 x 10?³ 4 | 0.075 | 0.075 | ? Using this information: (i) determine the rate equation for the reaction and the value of the rate constant; (ii) determine the expected rate of loss of NO for Experiment 4; (iii) from the reaction equation, show how the rate of loss of NO relates to the rate of loss of Cl?. Q1(b) [6 Marks]

Transcript

00:02 Okay, so here we want to determine the rate equation for this set of data that we have.

00:07 So first we know generically that the rate is going to be equal to k times the concentration of no raised to the x, and cl raised to the y.

00:22 So we can find the values for x and y based on the data that we're given here.

00:26 Here.

00:27 So first to find x, we want to set, we want to find two experiments where the chlorine remains constant but our nitric oxide is changing.

00:40 So let's look at experiments one and three.

00:44 So here for one and three, we see that we have 0 .1 to 1 .5 changing for nitric oxide.

00:53 So the way that we want to set this up is we take the concentration of experiment 3 divided by the concentration from experiment 1, we raise this to the value of x and this will be equal to the rate from experiment 3 divided by the rate from experiment 1.

01:14 And whatever rate or whatever amount or factor that the rate is changing should equal the left side which is the difference in the change in concentration raised to the power or the order of the reaction.

01:27 So here if we substitute in our values we get 0 .150 divided by 0 .100 raised to the x and we get 6 .75 times 10 to the negative 3 and 3 .00 times 10 to the negative 3.

01:45 So if we simplify we end up with 1 .5 raised to the x is equal to 2 .25.

01:53 So if we think about what our reaction orders typically are.

01:57 Typically it's going to be 0, 1, or 2.

02:00 So we can, if you're not quite sure what 1 .5 raised to the power of something is to be equal to 2 .25, you can start by substituting in these values, but we know if you raise something to a 0 exponent, it becomes 1.

02:16 If you raise something to a 1 exponent, the number stays the same.

02:19 So let's go ahead and try 2.

02:21 So if we square or 1 .5, it does end up being 2 .25.

02:29 So here our x is equal to 2.

02:32 Now we'll do the same thing for y.

02:34 We can look at experiments 1 and 2.

02:36 This is where nitric oxide stays constant, but chlorine changes.

02:41 And we'll set this up the same way.

02:43 So 0 .150 for chlorine from experiment 2, divided by 0 .100 from experiment 1, raised to the y, and we'll take our two rates.

02:53 So here we have 4 .5 times 10 to the negative 3 and 3 .00 times 10 to the negative 3...

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