Oxidation of bis(bipyridine) copper (I) ion by molecular oxygen is described by the equation:
Cu(C10H8N2)2+(aq) + O2(aq) ----products
The reaction is first order in oxygen and second order in Cu(C10H8N2)2+
(a) Write the rate law
(b) What is the overall reaction order?
(c) How does the reaction rate change if the concentration of Cu(C10H8N2)2+ is decreased by a factor of 4?
a.
rate = k(O2)[Cu(C10H8N2)2]^2
b.
2+1 = 3
c.
4^2 = 16 so decreasing by a factor of 4 would make it 1/16 the original rate.
(a) The rate law of the reaction can be written as:
Rate = k[O2][Cu(C10H8N2)2+]²
(b) The overall reaction order is the sum of the individual orders of the reactants in the rate law. In this case, the overall reaction order is 1 + 2 = 3.
(c) If the concentration of Cu(C10H8N2)2+ is decreased by a factor of 4, the reaction rate will decrease by a factor of 4² = 16.
(a) To write the rate law, we need to determine the individual reaction orders for both oxygen and Cu(C10H8N2)2+.
From the given information, we know that the reaction is first order in oxygen, meaning that the rate of the reaction is directly proportional to the concentration of oxygen: rate = k [O2]^1.
We also know that the reaction is second order in Cu(C10H8N2)2+. This means that the rate of the reaction is proportional to the square of the concentration of Cu(C10H8N2)2+: rate = k [Cu(C10H8N2)2+]^2.
Combining these, we can write the rate law for the reaction as:
rate = k [O2]^1 [Cu(C10H8N2)2+]^2
(b) The overall reaction order is the sum of the individual reaction orders. In this case, since the reaction is first order in oxygen and second order in Cu(C10H8N2)2+, the overall reaction order is 1 + 2 = 3.
(c) If the concentration of Cu(C10H8N2)2+ is decreased by a factor of 4, this means that its concentration becomes 1/4th of the initial concentration.
In the rate law, the concentration of Cu(C10H8N2)2+ is squared. Therefore, if the concentration of Cu(C10H8N2)2+ is decreased by a factor of 4, the overall rate of the reaction will decrease by a factor of (1/4)^2 = 1/16.
So, the reaction rate would decrease by a factor of 1/16 if the concentration of Cu(C10H8N2)2+ is decreased by a factor of 4.