Question

The progress of a reaction in the aqueous phase was monitered by the absorbance of a reactant at various times:

Time/s | 0 | 54 | 171 | 390 | 720 | 1010 | 1190 |

Absorbance | 1.67 | 1.51 | 1.24 | 0.847 | 0.478 | 0.301 | 0.216 |

Determine the order of the reaction and the rate constant.

Answer #1

Here we will consider that the fact that

Absrobance is directly proportional to concentration

So the variation of absrobance with time is similar to vairation of concentration with time

Now we will plot graphs so that we should obtain a straight line

Time | Absrobance | ln[A] |

0 | 1.67 | 0.512824 |

54 | 1.51 | 0.41211 |

171 | 1.24 | 0.215111 |

390 | 0.847 | -0.16605 |

720 | 0.478 | -0.73814 |

1010 | 0.301 | -1.20065 |

1190 | 0.216 | -1.53248 |

a) plot of [A] v/s time, if it is straight line then order of reaction is zero

so here is plot

This is not a straight line

b) let us draw a plot between ln[A] /s time, if it is a straight line then the reaction is first order

The straight line equation

y = -0.001x + 0.505 corresponds to ln[At] / [A0] = -kt

Therefore slope of the graph = -K

**K = 0.001 sec-1**

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