A. Based on the activation energies and frequency factors, rank the following reactions from fastest to slowest reaction rate, assuming they are all at the same temperature and that each starts with the same initial concentration.
E, 50 kJ/mol E,-350 kJ/mol 50 kJ/mol
A = 1.5 × 10-7 s-i A = 1.9 × 10-7 s-i A = 1.5 × 10-7 s-1
Fraction of molecules
The exponential term in the Arrhenius equation is equal to the fraction of molecules, f, with kinetic energy greater than or equal to the activation energy: f=e?Ea/(R?T). Most scientific calculators have an exfunction as the second function of the LN button.
B. A certain reaction with an activation energy of 165 kJ/mol was run at 505 K and again at 525 K . What is the ratio of f at the higher temperature to f at the lower temperature?

To answer this question we have to consider what factors affect reaction rate. To increase rate we need to have low activation energy (Ea) + high frequency factor (A). To decrease rate, we have high activation energy (Ea) + low frequency factor (A). Therefore, there's an inverse relationship between the two.

Part B:

First we need to calculate both f545 and f525. Formula for calculating both is the following: f = e^(-Ea/R*T)

Convert the 205 kJ/mol to Joules = 205000 = 2.05x10^5

f545 = e^(-2.05x10^5/8.314*545) = 2.25x10^-20

f525 = e^(-2.05x10^5/8.314*525) = 4.01x10^-21

Therefore, the final answer is f545/f525 = 2.25x10^-20/4.01x10^-21 = 5.610972569 = 6

oladayoademosu oladayoademosu · Answer 2 · 2022-03-30T22:16:33+02:00

a. Reactions

A = 1.9 × 10-7 s-1, E = -350 kJ/mol
A = 1.5 × 10-7 s-1, E = 50 kJ/mol
A = 1.5 × 10-7 s-1, E = 50 kJ/mol

b. Ratio of f at higher to lower temperature is 2.718

A. Arrhenius' equation

The three reactions are rated from fastest to slowest as

A = 1.9 × 10-7 s-1, E = -350 kJ/mol
A = 1.5 × 10-7 s-1, E = 50 kJ/mol
A = 1.5 × 10-7 s-1, E = 50 kJ/mol

The Arrhenius' equation gives the rate of reaction [tex]k = Ae^{-\frac{E_{a} }{RT} }[/tex] where

A = frequency factor,
Ea = activation energy,
R = molar gas constant and
T = temperature in kelvin

Now since each reaction starts at the same temperature, R and T are constant.

So, [tex]k = Ae^{-\frac{E_{a} }{RT} } \\k = Ae^{\frac{-E_{a}}{k}} \\k = Ae^{-{E_{a} }} e^{\frac{1}{k_{1} } } \\k = k_{2} Ae^{-{E_{a} }} ( k_{2} = e^{\frac{1}{k_{1} } })[/tex]

k ∝ A and k ∝ [tex]e^{-E_{a} }[/tex] ∝ 1/Ea

We see that the rate of reaction is directly proportional to the frequency factor and inversely proportional to the activation energy.

So,

for high rate of reaction we have high frequency factor and low activation energy and
for slow rate of reaction we have low frequency factor and high activation energy.

So, the three reactions are rated from fastest to slowest as

A = 1.9 × 10-7 s-1, E = -350 kJ/mol
A = 1.5 × 10-7 s-1, E = 50 kJ/mol
A = 1.5 × 10-7 s-1, E = 50 kJ/mol

B. Fraction of molecules

The ratio of f at the higher temperature to f at the lower temperature is 2.718

Since the fraction of molecules [tex]f = e^{-\frac{Ea}{RT} }[/tex] where

Ea = activation energy,
R = molar gas constant = 8.314 J/mol-K and
T = temperature in kelvin

Since the certain reaction with an activation energy of Ea = 165 kJ/mol was run at 505 K and again at 525 K,

Let f₁ = fraction of molecules at T₁ = 505 K, and f₂ = fraction of molecules at T₂ = 525 K

So, [tex]f_{1} = e^{-\frac{Ea}{RT_{1} } }[/tex] and [tex]f_{2} = e^{-\frac{Ea}{RT_{2} } }[/tex]

Ratio of f at the higher temperature to f at the lower temperature

So, the ratio of f at the higher temperature to f at the lower temperature is [tex]\frac{f_{2} }{f_{1} } = \frac{e^{-\frac{Ea}{RT_{2} } }}{e^{-\frac{Ea}{RT_{1} } }} \\\frac{f_{2} }{f_{1} } = e^{\frac{Ea}{RT_{1} } - \frac{Ea}{RT_{2} }} \\\frac{f_{2} }{f_{1} } = e^{\frac{E_{a} }{R} (\frac{1}{T_{1} } - \frac{1}{T_{2} })}[/tex]

Substituting the values of the variables into the equation, we have

[tex]\frac{f_{2} }{f_{1} } = e^{\frac{E_{a} }{R} (\frac{1}{T_{1} } - \frac{1}{T_{2} })}\\\frac{f_{2} }{f_{1} } = e^{\frac{165 000 J/mol}{8.314 J/mol-K} (\frac{1}{505 K} - \frac{1}{525K})}\\\frac{f_{2} }{f_{1} } = e^{{19846.043K}{ (0.00198/K - 0.001905/K)} }}\\\frac{f_{2} }{f_{1} } = e^{(19846.043K X 0.000075/K)}\\\frac{f_{2} }{f_{1} } = e^{1.4885}\\\frac{f_{2} }{f_{1} } = 2.718[/tex]

So, the ratio of f at the higher temperature to f at the lower temperature is 2.718

Learn more about Arrhenius' equation here:

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