Initial Rates Method: Find the Rate Equation

Initial rates, for I-B Chemistry H-L. This answer scores zero.

The student took the rate factor and called it the order. Can you see the mistake?

The order is a power. Two to the power of what gives four? Two. Second order in A.

That's one mark of an eight-mark Paper 2 style question. Seven more, and each one has a trap like that.

So where did that times four come from? And what's the order with respect to B?

Here's the data. Three experiments, at the same temperature. You compare two experiments where only one concentration changes. In experiments one and two, B stays the same. A doubles, and the rate goes from two point four to nine point six. Times four. That's where the four came from. And the same temperature matters, because k changes with temperature.

Now B. Experiments one and three. A stays the same, B triples, and the rate triples. So what's the order in B? Zero, one, or three?

One. Three to the power of one is three. First order in B. If the rate hadn't changed at all, that would be zero order. Any number to the power of zero is one. Mark two. So with both orders, what does the rate equation look like?

The orders become the powers. Rate equals k, times A squared, times B. Mark three. Notice it comes from your data, not from the balanced equation.

Now you find k. You rearrange, and you use experiment one. k equals the rate, over A squared times B.

Zero point zero one zero squared is one times ten to the minus four. Times zero point zero two zero, that's two times ten to the minus six. Two point four times ten to the minus four, over two times ten to the minus six. One hundred and twenty. Mark four.

You're the examiner now. Three students wrote the units of k. Only one of them gets the mark.

Answer A gave the units of a rate. No mark. Answer B gave the units for a second order reaction. No mark. Answer C, mol to the minus two, d-m to the six, per second. That one gets it.

Here's the quick way. Units of rate, divided by concentration, to the power of the overall order. Two plus one is three, so divide by concentration cubed.

mol d-m to the minus three per second, over mol cubed, d-m to the minus nine. For mol, one minus three is minus two. For d-m, minus three plus nine is six. And the per second stays. Mark five. So what rate do you predict for new concentrations?

Part c. A is zero point zero one five, and B is zero point zero four zero. Pause if you want to try it first; here it is.

One hundred and twenty, times zero point zero one five squared, times zero point zero four zero. One point zero eight times ten to the minus three. Mark six.

You can check it a second way. A is one and a half times experiment one. Squared, that's two point two five. B is doubled. Two point two five times two is four point five. Four point five times two point four times ten to the minus four. The same answer.

Part d is worth two marks. Why use initial rates at all? As the reaction runs, A and B get used up, so their concentrations keep falling. Only at the start do you know them exactly. They're the ones you mixed. Mark seven.

So how do you get an initial rate from a concentration time graph? A straight line from start to end, a tangent at time zero, or the concentration at time zero?

The tangent at time zero. A line from start to end gives an average rate, and the rate keeps falling. So you draw the tangent where the curve starts, and you find its gradient. Mark eight.

Concentration doubles, rate times four? Second order. Units of k, overall order three? mol to the minus two, d-m to the six, per second. The initial rate? The tangent at time zero.

What does a full-mark long answer look like? Next Saturday, a Biology Paper 2 question.