[student8607]

Rate constants are 1.3M^-1s^-1 at 700K & 23.0M^-1s^-1 at 800K.

I solved for the activation energy of 140kJ/mol and now have to determine the rate constant at 750K, so...

Ln (k₁/k₂) = -Ea/R x (1/T₂ - 1/T₁)
Ln(1.3/x) = -140,000J/8.31 x (1/750 - 1/700)
Ln(1.3/x) = 1.604
ANTI LN ANTI LN
1.3/x = 4.97/1
to get 0.2

but the answer in the book is 6.0?

[Astrokel]

Ln(1.3/x) = -140,000J/8.31 x (1/750 - 1/700)

[student8607]

Ln(1.3/x) = -140,000J/8.31 x (1/750 - 1/700)

Ln(1.3/x)=1.60x10¹⁰

But then how will I break up that fraction on the right.
Don't I need to do the anti-ln to cancel out the ln on the left side.

[Astrokel]

Ln(1.3/x)=1.60x1010

Your previous equation was fine except a mistake which i have already highlighted in red. So if you have adjust it, right side should be a negative value and your numerical value is way too off. And yes you need to do anti ln.

[student8607]

I did use the negative in my original post?

[Astrokel]

Yes the minus sign should not be present in your equation.

So it should read something like Ln(1.3/x) = 140,000J/8.31 x (1/750 - 1/700)

[student8607]

OHH. I see.

The handout from our teacher definitely has a negative sign in that equation.

Thanks!

[Astrokel]

k = Ae^-Ea/RT

k₁/k₂ = Ae^-Ea/RT₁/ Ae^-Ea/RT₂

k₁/k₂ = e^-Ea/RT₁/ e^-Ea/RT₂

k₁/k₂ = e^{-Ea/RT₁ - (-Ea/RT₂)}

k₁/k₂ = e^{-Ea/RT₁ + Ea/RT₂}

k₁/k₂ = e^{(Ea/R)(1/T₂ - 1/T₁)}

ln (k1/k2) = (Ea/R)(1/T₂ - 1/T₁)

Just a step-by-step to show you why there is no negative sign infront.

[student8607]

I see. Thanks. That helps.