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the activation energy ea for a particular reaction is 50.0 kj/mol. how much faster is the reaction at 313 k than at 310.0 k? (r

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To determine how much faster the reaction is at 313 K compared to 310.0 K, we can use the Arrhenius equation, which relates the rate constant (k) to the activation energy (Ea):

k = A * e^(-Ea/RT)

Where:
k is the rate constant
A is the pre-exponential factor (frequency factor)
Ea is the activation energy
R is the gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin

We can assume that the value of the pre-exponential factor (A) remains constant for the reaction.

Let's calculate the ratio of the rate constants at the two temperatures:

k1 = A * e^(-Ea/(R * T1))
k2 = A * e^(-Ea/(R * T2))

Divide k2 by k1 to get the ratio:

(k2 / k1) = (A * e^(-Ea/(R * T2))) / (A * e^(-Ea/(R * T1)))
= e^(-Ea/(R * T2) + Ea/(R * T1))
= e^(-Ea/R * (1/T2 - 1/T1))

Substitute the given values:
T1 = 310.0 K
T2 = 313 K
Ea = 50.0 kJ/mol (convert to J/mol)

Now we can calculate the ratio (k2 / k1) using the Arrhenius equation:

(k2 / k1) = e^(-Ea/R * (1/T2 - 1/T1))

Remember to convert the activation energy from kJ/mol to J/mol and to use the appropriate units for R.

By calculating this ratio, you will determine how much faster the reaction is at 313 K compared to 310.0 K.
User BalzacLeGeek
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