what is the value of the constant for a second order reaction if the reactant concentration drops from .657 M to ,0981 M in 17 s
1 answer:
Answer : The value of the constant for a second order reaction is,
Explanation :
The expression used for second order kinetics is:
where,
k = rate constant = ?
t = time = 17s
= final concentration = 0.0981 M
= initial concentration = 0.657 M
Now put all the given values in the above expression, we get:
Therefore, the value of the constant for a second order reaction is,
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Answer:
The volume of the final solution, V = 0.0305L
Explanation:
Number of moles = Concentration * volume
Concentration of HA = 1.00 * 10⁻⁴M
Volume of HA = 1000mL = 1 L
Number of moles of HA = 1.00 * 10⁻⁴ * 1
Number of moles of HA = 1.00 * 10⁻⁴ mols
Equation of reaction:
HA → H⁺ + A⁻
If 1 mol of HA produces 1 mol of H⁺ and A⁻, 1.00 * 10⁻⁴ mol of HA will produce 1.00 * 10⁻⁴ mol of H⁺ and A⁻.
Since only 16% dissociation occurs = 0.16
Number of moles of H⁺ produced = 0.16 * 1.00 * 10⁻⁴
Number of moles of H⁺ produced = 1.6 * 10⁻⁵mols
Number of moles of A⁻ produced = 0.16 * 1.00 * 10⁻⁴
Number of moles of A⁻ produced = 1.6 * 10⁻⁵mols
Since 16% of HA dissociated into H⁺ and A⁻, 84% of HA is left
Number of mols of HA left = 0.84 * 1.00 * 10⁻⁴
Number of mols of HA left = 8.4 * 10⁻⁵mols
Concentration = num of moles/volume
Let the volume of the final solution be V
Conc of HA = 8.4 * 10⁻⁵/V
Conc of H⁺ = 1.6 * 10⁻⁵/V
Conc of A⁻ = 1.6 * 10⁻⁵/V
To calculate the dissociation constant