Chemistry, 07.07.2019 06:00 yeehaw777

Acertain 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?

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Answer from: rainbowsauxe

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 from: pim9705876

$390 \; \text{K}$

Explanation

The rate of a chemical reaction is directly related to its rate constant 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

the frequency factor constant unique to this reaction

the base of natural logarithms, and

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,

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

Solve for T_2 :

$-(8.314 \times 10^{-3}/ 49.40) \; T_2 = 1/ (\ln 4.5 - 49.40 / (355 \times 8.314 \times 10^{-3})$

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

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