i was solving the specimen paper 2 page 14 rate of reaction table, and theres this one little doubt that i have, and because i dont get it, i lose 5 marks because the questions after are depending on it.

```

| Run | Initial concentration A (mol dm^-3) | Initial concentration B (mol dm^-3) | Initial concentration C (mol dm^-3) | Initial rate (mol dm^-3 s^-1) |

|-----|-------------------------------------|-------------------------------------|-------------------------------------|---------------------------------|

| 1 | 0.3 | 2 | 0 | 2.4 × 10^-3 |

| 2 | 0.6 | 4 | 0 | 4.8 × 10^-3 |

| 3 | 0.3 | 2 | 0.16 | 9.6 × 10^-3 |

| 4 | 0.6 | 4 | 0.3 | 0.23 |

```

i dont get how column C is first order, and not zero order??

```

| Run | Initial concentration A (mol dm^-3) | Initial concentration B (mol dm^-3) | Initial concentration C (mol dm^-3) | Initial rate (mol dm^-3 s^-1) |

|-----|-------------------------------------|-------------------------------------|-------------------------------------|---------------------------------|

| 1 | 0.3 | 2 | 0 | 2.4 × 10^-3 |

| 2 | 0.6 | 4 | 0 | 4.8 × 10^-3 |

| 3 | 0.3 | 2 | 0.16 | 9.6 × 10^-3 |

| 4 | 0.6 | 4 | 0.3 | 0.23 |

```

i dont get how column C is first order, and not zero order??

(edited 1 month ago)

Original post by ly2021

i was solving the specimen paper 2 page 14 rate of reaction table, and theres this one little doubt that i have, and because i dont get it, i lose 5 marks because the questions after are depending on it.

```

| Run | Initial concentration A (mol dm^-3) | Initial concentration B (mol dm^-3) | Initial concentration C (mol dm^-3) | Initial rate (mol dm^-3 s^-1) |

|-----|-------------------------------------|-------------------------------------|-------------------------------------|---------------------------------|

| 1 | 0.3 | 2 | 0 | 2.4 × 10^-3 |

| 2 | 0.6 | 4 | 0 | 4.8 × 10^-3 |

| 3 | 0.3 | 2 | 0.16 | 9.6 × 10^-3 |

| 4 | 0.6 | 4 | 0.3 | 0.23 |

```

i dont get how column C is first order, and not zero order??

```

| Run | Initial concentration A (mol dm^-3) | Initial concentration B (mol dm^-3) | Initial concentration C (mol dm^-3) | Initial rate (mol dm^-3 s^-1) |

|-----|-------------------------------------|-------------------------------------|-------------------------------------|---------------------------------|

| 1 | 0.3 | 2 | 0 | 2.4 × 10^-3 |

| 2 | 0.6 | 4 | 0 | 4.8 × 10^-3 |

| 3 | 0.3 | 2 | 0.16 | 9.6 × 10^-3 |

| 4 | 0.6 | 4 | 0.3 | 0.23 |

```

i dont get how column C is first order, and not zero order??

First you know that A is first order: given row 1 and 4 the scale factor for initial concentration and initial rate are equal: scale factor of x2.

For B: row 1 and 3 are used since only B changes but notice they have diff scale factors of 2 and 4 respectively for initial conc and initial rate. Hence 2 is squared.

For C we can solve with row 1 and 4 knowing orders of A and B. A = doubles the conc and does the same to rate. So firstly rate inc by scale factor of 2. B inc by scale factor of x4 so this will inc rate by x 16. So the rate is currently at 0.0768 given these scale factors. Now you see that the concentration for C inc by a factor of x3. Initial rate of column 4 is 0.23. From 0.0768 to 0.23 there has been an inc by scale factor x3. Concentration scale factor and rate scale factor both x3 so it's first order. BTW I think some of your values above are wrong.

Assuming I found the correct paper:

1) You should have worked out that the reaction is first order with respect to A and second order with respect to B

2) To find the order with respect to C, you can compare any run with run 4, now that you know the order with respect to A and B. Let's take run 3 as an example. [A] doubles, so you can multiply inital rate by 2, knowing that A is first order. also doubles, so you can further multiply the inital rate by 2^{2}, as B is second order. When you divide the inital rate of run 4 by this value calculated, you should get 3. As [C] has tripled, this tells you that the reaction is first order with respect to C.

1) You should have worked out that the reaction is first order with respect to A and second order with respect to B

2) To find the order with respect to C, you can compare any run with run 4, now that you know the order with respect to A and B. Let's take run 3 as an example. [A] doubles, so you can multiply inital rate by 2, knowing that A is first order. also doubles, so you can further multiply the inital rate by 2

Original post by Methene

Assuming I found the correct paper:

1) You should have worked out that the reaction is first order with respect to A and second order with respect to B

2) To find the order with respect to C, you can compare any run with run 4, now that you know the order with respect to A and B. Let's take run 3 as an example. [A] doubles, so you can multiply inital rate by 2, knowing that A is first order. also doubles, so you can further multiply the inital rate by 2^{2}, as B is second order. When you divide the inital rate of run 4 by this value calculated, you should get 3. As [C] has tripled, this tells you that the reaction is first order with respect to C.

1) You should have worked out that the reaction is first order with respect to A and second order with respect to B

2) To find the order with respect to C, you can compare any run with run 4, now that you know the order with respect to A and B. Let's take run 3 as an example. [A] doubles, so you can multiply inital rate by 2, knowing that A is first order. also doubles, so you can further multiply the inital rate by 2

so initially when i do 0.48/0.16 = 3.

so how would it mean its first order, when it has tripled

Original post by Tulipbloom

First you know that A is first order: given row 1 and 4 the scale factor for initial concentration and initial rate are equal: scale factor of x2.

For B: row 1 and 3 are used since only B changes but notice they have diff scale factors of 2 and 4 respectively for initial conc and initial rate. Hence 2 is squared.

For C we can solve with row 1 and 4 knowing orders of A and B. A = doubles the conc and does the same to rate. So firstly rate inc by scale factor of 2. B inc by scale factor of x4 so this will inc rate by x 16. So the rate is currently at 0.0768 given these scale factors. Now you see that the concentration for C inc by a factor of x3. Initial rate of column 4 is 0.23. From 0.0768 to 0.23 there has been an inc by scale factor x3. Concentration scale factor and rate scale factor both x3 so it's first order. BTW I think some of your values above are wrong.

For B: row 1 and 3 are used since only B changes but notice they have diff scale factors of 2 and 4 respectively for initial conc and initial rate. Hence 2 is squared.

For C we can solve with row 1 and 4 knowing orders of A and B. A = doubles the conc and does the same to rate. So firstly rate inc by scale factor of 2. B inc by scale factor of x4 so this will inc rate by x 16. So the rate is currently at 0.0768 given these scale factors. Now you see that the concentration for C inc by a factor of x3. Initial rate of column 4 is 0.23. From 0.0768 to 0.23 there has been an inc by scale factor x3. Concentration scale factor and rate scale factor both x3 so it's first order. BTW I think some of your values above are wrong.

Original post by ly2021

so initially when i do 0.48/0.16 = 3.

so how would it mean its first order, when it has tripled

so how would it mean its first order, when it has tripled

The concentration of C has tripled, and the inital rate has tripled (after taking into account A and B). If C was second order, tripling the concentration of C would increase the inital rate by a factor of 3

Original post by Methene

The concentration of C has tripled, and the inital rate has tripled (after taking into account A and B). If C was second order, tripling the concentration of C would increase the inital rate by a factor of 3^{2} . If C was zero order, tripling the concentration of C would have no effect on the inital rate.

sorry if im asking too much, but which rate value do we look at when we are considering the idea that the initial rate has also tripled, because for example if im taking 0.0024 and triple it its 0.0072, if you see where im coming from. so initially which value will we consider when multiplying by 3

Original post by ly2021

sorry if im asking too much, but which rate value do we look at when we are considering the idea that the initial rate has also tripled, because for example if im taking 0.0024 and triple it its 0.0072, if you see where im coming from. so initially which value will we consider when multiplying by 3

No worries, I wasn't very clear. Let's take run 3 inital rate and compare to run 4. Run 3 inital rate is 0.0096 and for run 4 it's 0.23. The concentration of A has doubled. A is first order, so we can multiply 0.0096 x 2 = 0.0192. The concentration of B has also doubled. B is second order, so we can multiply 0.0192 by 2

Now if we triple this value, we end up with the inital rate for run 4 (0.23). We can also see that the concentration of C has tripled for run 4, compared to run 3. Therefore C must be first order.

This might help to make it easier to understand:

The rate equation for the reaction would be rate = k [A]

If we doubled the concentration of A, the rate would double. Rate = k (2 x [A])

If we doubled the concentration of B, the rate would quadruple. Rate = k [A] (2 x )

From this logic, if [C] has tripled and the inital rate has also tripled (after accounting for A and B), then C must be first order.

Original post by Methene

No worries, I wasn't very clear. Let's take run 3 inital rate and compare to run 4. Run 3 inital rate is 0.0096 and for run 4 it's 0.23. The concentration of A has doubled. A is first order, so we can multiply 0.0096 x 2 = 0.0192. The concentration of B has also doubled. B is second order, so we can multiply 0.0192 by 2^{2} = 0.0768.

Now if we triple this value, we end up with the inital rate for run 4 (0.23). We can also see that the concentration of C has tripled for run 4, compared to run 3. Therefore C must be first order.

This might help to make it easier to understand:

The rate equation for the reaction would be rate = k [A]^{2} [C]

If we doubled the concentration of A, the rate would double. Rate = k (2 x [A])^{2 }[C]

If we doubled the concentration of B, the rate would quadruple. Rate = k [A] (2 x )^{2} [C]

From this logic, if [C] has tripled and the inital rate has also tripled (after accounting for A and B), then C must be first order.

Now if we triple this value, we end up with the inital rate for run 4 (0.23). We can also see that the concentration of C has tripled for run 4, compared to run 3. Therefore C must be first order.

This might help to make it easier to understand:

The rate equation for the reaction would be rate = k [A]

If we doubled the concentration of A, the rate would double. Rate = k (2 x [A])

If we doubled the concentration of B, the rate would quadruple. Rate = k [A] (2 x )

From this logic, if [C] has tripled and the inital rate has also tripled (after accounting for A and B), then C must be first order.

THANK U SMM!!! it makes so much more sense now, i appreciate your help so much xxx

Original post by ly2021

THANK U SMM!!! it makes so much more sense now, i appreciate your help so much xxx

You're welcome 😁

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