
For the reaction \[{[Cu{(N{H_3})_4}]^{2 + }} + {H_2}O \to {[Cu{(N{H_3})_3}{H_2}O]^{2 + }} + N{H_3}\]
The net rate of reaction at any time is given by
Net rate= \[ - 2.0 \times {10^{ - 4}}{[Cu{(N{H_3})_4}]^{2 + }} = 3.0 \times {10^5}{[Cu{(N{H_3})_3}{H_2}O]^{2 + }}[N{H_3}]\]
Then correct statement is (are):
A. Rate constant for forward reaction = \[2 \times {10^{ - 4}}\]
B. Rate constant for backward reaction = \[3 \times {10^5}\]
C. Equilibrium constant for the reaction= \[6.6 \times {10^{ - 10}}\]
D. All of the above
Answer
566.7k+ views
Hint: We have to know that the rate of reaction is the pace or speed at which a chemical reaction occurs. Reaction rate means that the speed at which reactants are converted into their products. Reaction rates can vary a lot. The forward and the backward reaction also play an important role.
Formula used: constant=kb/kt
kb is the base dissociation constant
kT is the amount of heat required to increase entropy in a thermodynamic system
Complete step by step answer:
Given data contains,
Net rate=\[{k_b}{[Cu{(N{H_3})_3}{H_2}O]^{2 + }} + [N{H_3}]\]
kt =\[2 \times {10^{ - 4}}\]
kb=\[3 \times {10^5}\]
Constant= \[\dfrac{{{k_t}}}{{{k_b}}} = \dfrac{{2 \times {{10}^{ - 4}}}}{{3 \times {{10}^5}}} = 6.6 \times {10^{ - 10}}\]
So option D is correct which is that all the above statements are correct.
Which is that the forward reaction is \[2.0 \times {10^{ - 4}}\] and the backward reaction is \[3.0 \times {10^5}\] and the equilibrium constant of the reaction is \[6.6 \times {10^{ - 10}}\] .
So we can conclude that all the statements are correct. So we can also say that option A , B, C are not valid as they only have a small amount of information.
All of them are correct hence option D is the correct one.
Note:
We know that the Kt reaction can also be called as the forward reaction and kb reaction can also be called as the backward reaction. Both the forward and backward reactions play a very vital role in the solution of the given problem. All the given statements are correct as all of them are true. Kt is a product of Boltzmann constant, and the temperature, T. kT is also the amount of heat required for increasing the thermodynamic entropy of a system. Kb is base dissociation constant. The base dissociation constant helps to know how completely a base dissociates in water, in the dissociation it we find component ions.
Formula used: constant=kb/kt
kb is the base dissociation constant
kT is the amount of heat required to increase entropy in a thermodynamic system
Complete step by step answer:
Given data contains,
Net rate=\[{k_b}{[Cu{(N{H_3})_3}{H_2}O]^{2 + }} + [N{H_3}]\]
kt =\[2 \times {10^{ - 4}}\]
kb=\[3 \times {10^5}\]
Constant= \[\dfrac{{{k_t}}}{{{k_b}}} = \dfrac{{2 \times {{10}^{ - 4}}}}{{3 \times {{10}^5}}} = 6.6 \times {10^{ - 10}}\]
So option D is correct which is that all the above statements are correct.
Which is that the forward reaction is \[2.0 \times {10^{ - 4}}\] and the backward reaction is \[3.0 \times {10^5}\] and the equilibrium constant of the reaction is \[6.6 \times {10^{ - 10}}\] .
So we can conclude that all the statements are correct. So we can also say that option A , B, C are not valid as they only have a small amount of information.
All of them are correct hence option D is the correct one.
Note:
We know that the Kt reaction can also be called as the forward reaction and kb reaction can also be called as the backward reaction. Both the forward and backward reactions play a very vital role in the solution of the given problem. All the given statements are correct as all of them are true. Kt is a product of Boltzmann constant, and the temperature, T. kT is also the amount of heat required for increasing the thermodynamic entropy of a system. Kb is base dissociation constant. The base dissociation constant helps to know how completely a base dissociates in water, in the dissociation it we find component ions.
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