#### Question

From the following data for the liquid phase reaction A → B, determine the order of reaction

and calculate its rate constant

t/s | 0 | 600 | 1200 | 1800 |

[A]/`molL^-1` | 0.624 | 0.446 | 0.318 | 0.226 |

#### Solution

Calculation:

We require to calculate the rate constant at different time intervals.

`k=(2.303)/t log_10[ [A]_0/[A]_t]` ...........[Integrated rate law for first order reaction]

**A.**

`[A]_0=0.624`

`[A]_(t1)=0.446`

t1=600s

`k_1=2.303/t_1 log[[A]_0/[A]_t]`

`k_1=2.303/600 log [0.624/0.446]`

`k_1=3.838xx10^-3 log[1.399]`

`k_1=3.838xx10^-3xx0.1458`

`therefore k_1=5.596xx10^-4 s^-1`

**B.**

`[A]_0=0.624`

`[A]_t=0.318`

`t_2=1200s`

`k_2=2.303/t_2 log_10[0.624/0.318]`

`k_2=1.919xx10^-3 log(1.962)`

`k_2=1.919xx10^3xx0.2927`

`k_2=5.617xx10^-4 s^-1`

**C.**

`[A]_0=0.624`

`[A]_(t3)=0.226`

`t_3=1800 s`

`k_3=2.303/1800 log[0.624/0.226]`

`k_3=1.279xx10^-3xx0.4411`

`k_3=5.641xx10^-4s^-1`

`k=(k_1+k_2+k_3)/3=((5.596xx10^-4)+(5.617xx10^-4)+(5.641xx10^-4))/3`

`k=5.618xx10^-4 s^-1`

All the k values calculated at different time intervals are almost the same. This implies that the reaction obeys the integrated rate equation of first order reaction. Hence, the reaction is a first order reaction.