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rates and orders confusion

im really confused by orders in Module 4 of chemistry. I dont get how you work out if something is first order, second etc. when given concentrations and rates such as

Initial conc. of A 0.15, 0.30, 0.60
Initial conc. of B 0.24, 0.24, 0.48
Initial rate 0.45 x 10^-5, 0.90 x 10^-5, 7.20 x 10^-5

can someone tell me what the order with respect to A and the order with respect to B is and how they worked it out please. My teacher didnt explain it very well

Reply 1

modini

sharp357

A is first order. B is second order.

For A: The rate increases by a factor of 2 when the concentration of A increases by a factor of 2 (directly proportional).

So: when stays the same and [A] gets increased from 0.15 to 0.3, the initial rate increases from 0.45 x 10^-5 to 0.9 x 10^-5

For B: The rate increases by a factor of squared. So when they increase by a factor of 2, the initial rate increases by a factor of 2^2.

When [A] increased from 0.30 to 0.60 and increased from 0.24 to 0.48, the initial rate goes from 0.90 x 10^-5 to 7.20 x 10^-5. We know A is first order, so increasing [A] from 0.30 to 0.60 will increase the initial rate from 0.9 x 10^-5 to 1.8 x 10^-5.

The new initial rate is 7.2 x 10^-5.

7.2 x 10^-5/1.8 x 10^-5 = 4 which is 2^2 hence second order with respect to B.

Hope that's a bit clearer for you :biggrin: