A researcher raises the temperature from 46.5 to 64.4°C and finds that the rate of the reaction doubles. What was the activation energy (in J/mol) for this reaction? (R = 8.3145 J/mol·K)

The activation energy of a reaction, which explains the effect of temperature on reaction rate, can be calculated using the rearranged form of the Arrhenius equation. When the temperature was increased from 46.5 to 64.4 degrees Celsius and the reaction rate doubled, it allowed for the calculation of the activation energy in joules per mole.

Explanation:

The subject of the question pertains to the concept of

activation energy

in Chemistry, particularly its determination based on the effect of temperature on reaction rate according to the Arrhenius Equation. Given the observed increase in reaction rate when temperature is raised from 46.5 to 64.4 degree Celsius, the activation energy can be calculated using the rearranged form of the Arrhenius Equation: ln(k2/k1) = (Ea/R)*(1/T1 - 1/T2) where k1 and k2 are the rate constants at T1(T1=273.15+46.5=319.65 K) and T2 (T2=273.15+64.4=337.55 K), respectively. Given that the rate of reaction doubled, this suggests k2 is 2*k1.

R

is the ideal gas constant (8.3145 J/mol/K),

Ea

is the activation energy. Solving this equation would give the activation energy of reaction in J/mol.

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