## 7D.1

Arrhenius Equation:

Jason Wu 1E
Posts: 101
Joined: Thu Jul 25, 2019 12:15 am

### 7D.1

The rate constant of the first-order reaction 2 N2O(g) -> 2 N2(g) + O2(g) is 0.76 s^-1 at 1000. K and 0.87 s^-1 at 1030. K. Calculate the activation energy of the reaction.

How do we approach this question? are we supposed to substitute for k and T into Arrhenius's Equation to create a system of equations to solve for A and Ea? How do we do this calculation?

Katherine Wu 1H
Posts: 104
Joined: Fri Aug 30, 2019 12:15 am
Been upvoted: 2 times

### Re: 7D.1

we use ln(k prime/k) = (Ea/R)(1/T-1/T prime) = (Ea/R)((T prime-T)/T(T prime))
ln (k prime/k)=ln(0.87 s^-1/0.76 s^-1)
=(Ea/8.31x10^-3 kJ/K.mol)((1030K-1000K)/(1030Kx1000K))
Ea=((8.31x10^-3kJ/K.mol)(1000K)(1030K)/(1030K-1000K))ln(0.87 s^-1/0.76 s^-1)
=39 kJ/mol

Shannon Asay 1C
Posts: 102
Joined: Fri Aug 09, 2019 12:16 am

### Re: 7D.1

I believe you could use the equation ln(k2/k1)=-Ea/R*(1/T2-1/T1). Just plug in all your variables and solve for Ea.

Jessa Maheras 4F
Posts: 121
Joined: Fri Aug 02, 2019 12:16 am

### Re: 7D.1

Shannon Asay 1C wrote:I believe you could use the equation ln(k2/k1)=-Ea/R*(1/T2-1/T1). Just plug in all your variables and solve for Ea.

Yes, I agree! All you do is rearrange the equation and plug in values.

Aman Sankineni 2L
Posts: 103
Joined: Fri Aug 30, 2019 12:17 am

### Re: 7D.1

The equation ln(k2/k1)=-Ea/R*(1/T2-1/T1) should be able to be used.