Lyndon #13c


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davidbakalov_lec2_2L
Posts: 61
Joined: Fri Sep 28, 2018 12:23 am

Lyndon #13c

Postby davidbakalov_lec2_2L » Sat Mar 16, 2019 6:30 pm

Can someone explain 13c from Lyndon's review sheet?

Alana Sur 3B
Posts: 61
Joined: Fri Sep 28, 2018 12:25 am

Re: Lyndon #13c

Postby Alana Sur 3B » Sat Mar 16, 2019 8:27 pm

Pre exponential factor is A and you use the arrhenius equation to solve for it : k = A x e^(-Ea/RT)

Nicole Elhosni 2I
Posts: 62
Joined: Fri Sep 28, 2018 12:28 am

Re: Lyndon #13c

Postby Nicole Elhosni 2I » Sat Mar 16, 2019 8:31 pm

We use the equation lnk= -Ea/RT + lnA, and we are solving for A.
You get k from the question above, so ln(50.45 M⁻²s⁻¹)=-(23 x 10³ J/mol)/(8.314 J/Kmol x 298K) +lnA
lnA=ln(50.45 M⁻²s⁻¹) + (23 x 10³ J/mol)/(8.314 J/Kmol x 298K)
So A=5.43x10⁵ M⁻²s⁻¹.


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