Question

Part A

The activation energy of a certain reaction is 40.6 kJ/mol . At 20 ∘C , the rate constant is 0.0150s−1 . At what temperature in degrees Celsius would this reaction go twice as fast?

Express your answer with the appropriate units.

Part B

Given that the initial rate constant is 0.0150s−1 at an initial temperature of 20 ∘C , what would the rate constant be at a temperature of 110. ∘C for the same reaction described in Part A?

Express your answer with the appropriate units.

± The Arrhenius Equation

The Arrhenius equation shows the relationship between the rate constant k and the temperature T in kelvins and is typically written as

k=Ae−Ea/RT

where R is the gas constant (8.314 J/mol⋅K ), A is a constant called the frequency factor, and Ea is the activation energy for the reaction.

However, a more practical form of this equation is

lnk2k1=EaR(1T1−1T2)

which is mathmatically equivalent to

lnk1k2=EaR(1T2−1T1)

where k1 and k2 are the rate constants for a single reaction at two different absolute temperatures (T1 and T2 ).

Answer 1

A)

if we want 2x (twice) as fast, then rate must be twice, therefore, rate constant must be 2x that of the original

K2 = 2*K1

K1 = K1, no need ot use numbers

Ahrrenius Equation 2 Points

From Ahrrenius equation;

K1 = A*exp(-Ea/(RT1))

K2 = A*exp(-Ea/(RT2))

Note that A and Ea are the same, they do not depend on Temperature ( in the range fo temperature given)

Then

Divide 2 and 1

K2/K1 = A/A*exp(-Ea/(RT2)) / exp(-Ea/(RT1))

Linearize:

ln(K2/K1) = -Ea/R*(1/T2-1/T1)

get rid of negative sign

ln(K2/K1) = Ea/R*(1/T1-1/T2)

now, substitute data known

ln(2K1/K1) = 40600/8.314*(1/(293) - 1/(T2))

note that T2 is in (K)

solve

ln(2) =  40600/8.314*(1/(293) - 1/(T2))

ln(2)*8.314/40600 - 1/293 = -1/T2

T2 = 0.0032710^-1

T2 = 0.0032710 =O 305.7169 K

T2 = 305.7169-273 = 32.7169 C

T = 32.7169 C must be in order to have a faster rate

B)

using the same equation:

ln(K2/K1) = Ea/R*(1/T1-1/T2)

now we need to solve for K2

Substitute all data but K2

ln(K2/(0.015)) = 40600/8.314*(1/(293) - 1/(110+273))

K2 = 0.015*exp(3.91644)

K2 = 0.753320 1/s