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 :

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

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

Answer from: Quest

Abbreviated as mg-25.

Answer from: Quest

for step 7 table c on the lab "newtons law of action" the average time it took the car to travel .25 meters was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

hope this : )

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Acertain reaction has an activation energy of 49.40 kj/mol. at what kelvin temperature will the reac...