In the mirror diagram shown, Which is the normal?
2 answers:
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Answer:
B. First order, Order with respect to C = 1
Explanation:
The given kinetic data is as follows:
A + B + C → Products
[A]₀ [B]₀ [C]₀ Initial Rate (10⁻³ M/s)
1. 0.4 0.4 0.2 160
2. 0.2 0.4 0.4 80
3. 0.6 0.1 0.2 15
4. 0.2 0.1 0.2 5
5. 0.2 0.2 0.4 20
The rate of the above reaction is given as:
where x, y and z are the order with respect to A, B and C respectively.
k = rate constant
[A], [B], [C] are the concentrations
In the method of initial rates, the given reaction is run multiple times. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.
Order w.r.t A : Use trials 3 and 4
Order w.r.t B : Use trials 2 and 5
Order w.r.t C : Use trials 1 and 2
we know that x = 1 and y = 2, substituting the appropriate values in the above equation gives:
z = 1
Therefore, order w.r.t C = 1
Through manipulation of equations, we are able to obtain the equation:
![-pOH= log [ OH^{-}]](https://tex.z-dn.net/?f=-pOH%3D%20log%20%5B%20OH%5E%7B-%7D%5D%20)
Then we can transform the equation into:
![[ OH^{-}]= 10^{-pOH}](https://tex.z-dn.net/?f=%5B%20OH%5E%7B-%7D%5D%3D%2010%5E%7B-pOH%7D%20%20)
Then we are able to plug in the pOH and directly get [OH-]:
![[ OH^{-}] = 10^{-6.48}](https://tex.z-dn.net/?f=%5B%20OH%5E%7B-%7D%5D%20%3D%2010%5E%7B-6.48%7D%20)
![[ OH^{-}]=3.31* 10^{-7} M](https://tex.z-dn.net/?f=%5B%20OH%5E%7B-%7D%5D%3D3.31%2A%2010%5E%7B-7%7D%20M%20%20)
Then we can transform the equation into:
Then we are able to plug in the pOH and directly get [OH-]: