Question

A certain reaction has an activation energy of 49.40 kJ/mol. At what Kelvin temperature will the reaction proceed 4.50 times faster than it did at 355 K?

Answer 1

Given:

Ea(Activationenergy):49.4kJ/molecule

k1: the rate constant of the first reaction

k2 : rate constant of the second reaction.

T2: Temperature of the second reaction.

T1: Temperature of the first reaction.

k2/k1=4.55

Now by Arrhenius equation we get

log(k2/k1)=[Ea/(2.303xR)] x[(1/T1)-(1/T2)]

Where k1 is the rate constant of the first reaction.

k2 is the rate constant of the second equation.

T2 is the temperature of the second reaction measured in K

T1 is the temperature of the first reaction measured in K

Ea is the activation energy kJ/mol

R is the gas constant measured in J/mol.K

Now substituting the given values in the Arrhenius equation we get:

log(k2/k1)=[Ea/(2.303xR)] x[(1/T1)-(1/T2)]

log(4.55)=[Ea/(2.303xR)] x[(1/T1)-(1/T2)]

0.66=[49.4/(2.303x8.314x10^-3)]x[(1/355)-(1/T2)]

0.66= 2579.75x [(1/355)-(1/T2)]

0.000256= (T2-355)/355T2

0.0908T2-T2= -355

0.9092T2=355

T2=390.46K

Answer 2

`390 \; \text{K}`

Step-by-step explanation

The rate of a chemical reaction is directly related to its rate constant
`k` if the concentration of all reactants in its rate-determining step is held constant. The rate "constant" is dependent on both the temperature and the activation energy of this particular reaction, as seen in the Arrhenius equation:

`k = A \cdot e^{-E_A/( R\cdot T)`

where

• 
  `A` the frequency factor constant unique to this reaction
• 
  `e` the base of natural logarithms, and
• 
  `R` the ideal gas constant.

Taking natural logarithms of both sides of the expression yields:

`\ln k = \ln A - {E_a}/ ({R \cdot T})`

`k_2 = 4.50 \; k_1`, such that

`\ln k_2 = \ln 4.5 + \ln k_1`

`\ln A- {E_a}/ ({R \cdot T_2}) = \ln k_2 \\\phantom{\ln A-{E_a}/ ({R \cdot T_2})} = \ln 4.5 + \ln k_1\\ \phantom{\ln A- {E_a}/ ({R \cdot T_2})} = \ln 4.5 +\ln A- {E_a}/ ({R \cdot T_1})`

Rearranging gives

`-{E_a}/ ({R \cdot T_2}) = \ln 4.5- {E_a}/ ({R \cdot T_1})`

Given the initial temperature
`T_1 = 355 \; \text{K}` and activation energy
`E_A = 49.40 \; \text{kJ} \cdot \text{mol}^(-1)`- assumed to be independent of temperature variations,

`- {49.40 \; \text{kJ} \cdot \text{mol}^(-1)}/ ({8.314 * 10^(-3) \; \text{kJ} \cdot \text{mol}^(-1) \cdot \text{K}^(-1) \cdot T_2}) \\= \ln 4.5- {49.40 \; \text{kJ} \cdot \text{mol}^(-1)}/ ({355 \; \text{K}\cdot 8.314 * 10^(-3) \; \text{kJ} \cdot \text{mol}^(-1) \cdot \text{K}^(-1)})`

Solve for
`T_2`:

`-(8.314 * 10^(-3)/ 49.40) \; T_2 = 1/ (\ln 4.5 - 49.40 / (355 * 8.314 * 10^(-3))`

`T_2 = 390 \; \text{K}`