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xeze [42]
3 years ago
11

A certain reaction with an activation energy of 185 kJ/mol was run at 505 K and again at 525 K . What is the ratio of f at the h

igher temperature to f at the lower temperature
Chemistry
1 answer:
frosja888 [35]3 years ago
5 0

Answer:

The ratio of f at the higher temperature to f at the lower temperature is 5.356

Explanation:

Given;

activation energy, Ea = 185 kJ/mol = 185,000 J/mol

final temperature, T₂ = 525 K

initial temperature, T₁ = 505 k

Apply Arrhenius equation;

Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]

Where;

\frac{f_2}{f_1}  is the ratio of f at the higher temperature to f at the lower temperature

R is gas constant = 8.314 J/mole.K

Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]\\\\Log(\frac{f_2}{f_1} ) = \frac{185,000}{2.303 \times 8.314} [\frac{1}{505} -\frac{1}{525} ]\\\\Log(\frac{f_2}{f_1} ) = 0.7289\\\\\frac{f_2}{f_1}  = 10^{0.7289}\\\\\frac{f_2}{f_1}  = 5.356

Therefore, the ratio of f at the higher temperature to f at the lower temperature is 5.356

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