Multiple Questions

Question

ch 17 q 23

A certain reaction has an activation energy of 36.97 kJ/mol.36.97 kJ/mol. 

At what Kelvin temperature will the reaction proceed 6.506.50 times faster than it did at 357 K?

ch 17 q36

A certain reaction has an enthalpy of Δ𝐻=−44 kJΔ⁢H=−44 kJ and an activation energy of 𝐸a=53 kJ.Ea=53 kJ.

What is the activation energy of the reverse reaction?

𝐸a(reverse)=______kJ

ch17 q14

The rate constant for this second‑order reaction is 0.460 M−1⋅s−10.460⁢ M−1⋅s−1 at 300 ∘C.300⁢ ∘C.

A⟶productsA⟶products

How long, in seconds, would it take for the concentration of AA to decrease from 0.820 M0.820 M to 0.350 M?

t=__s

Answer 1

Tatum,

You are asking a lot of questions and it is doubtful anyone will answer all of them. We don't do homework for people, and if you attempt to solve the problem(s), and present your effort, then maybe will get more assistance. I will attempt to "get you started", but hope you'll be able to do the rest.

q23: ln(k2/k1) = -Ea/R (1/T2 - 1/T1)

k2/k1 = 6.50

Ea = 36.97 kJ/mol

R = 0.008314 kJ/molK

T1 = 357K

T2 = ?

Solve for T2

q36: draw an energy curve with products lower than reactants, and Ea hill = 53. The energy of the products will be 44 kJ below that of reactants. Thus the reverse reaction will have Ea = 44 kJ + 43 kJ = 97 kJ

q14: I answered this in another of you questions. Please refer to that post and please stop asking the same question over again.