What did Bohr’s model of the atom do that Rutherford’s model did not?
2 answers:
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The question is incomplete, complete question is :
The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25°C.
Activation energy of the reaction 1 , = 14.0 kJ/mol
Activation energy of the reaction 2, = 11.9 kJ/mol
Answer:
0.4284 is the ratio of the rate constants.
Explanation:
According to the Arrhenius equation,
The expression used with catalyst and without catalyst is,
where,
= rate constant reaction -1
= rate constant reaction -2
Activation energy of the reaction 1 , = 14.0 kJ/mol = 14,000 J
Activation energy of the reaction 2, = 11.9 kJ/mol = 11,900 J
R = gas constant = 8.314 J/ mol K
T = temperature =
Now put all the given values in this formula, we get
0.4284 is the ratio of the rate constants.
"cg" is centigram, which is one-hundredth of a gram.
I will first convert from g to cg (multiply by 100), then from mL to L (multiply by 1000).