# The Ea is 93.1 kJ/mol. At 600 K, the rate constant is 3.37 × 103 M-1 s-1. At what temperature will the rate constant be 27.0 × 103 M-1 s-1?

Ozone decomposes to form oxygen molecules and oxygen atoms, O3(g) → O2(g) + O(g), in the upper atmosphere. The energy of activation for this reaction is 93.1 kJ/mol. At 600 K, the rate constant for this reaction is 3.37 × 103 M-1 s-1. At what temperature will the rate constant be 8 times larger, or 27.0 × 103 M-1 s-1?

I've tried this problem before and have gotten multiple answers. The one I think may be correct is 539.8119 K, but the homework system is not taking my answer. I'm just wondering if I did it wrong.

The equation i used is:

ln(k2/k1)=(-Ka/8.314)(1/T2-1/T1)

### 1 Answer

- hcbiochemLv 73 weeks agoFavourite answer
That is the right equation to use. You can also write it as:

ln (k2/k1) = Ea/R (1/T1 - 1/T2)

This is obtained by multiplying the terms in the parenthesis by -1. So,

ln (8) = 93100 J/mol/8.314 J/molK (1/600K - 1/T2)

T2 = 675 K

I'm not sure how you are getting the answer you are. Also, your answer can have only 3 significant figures. All the digits after the decimal point are meaningless.