I know the enthalpy of a reaction is 23kj/mol. Initially the reaction is taking place at 273 k. To what temperature do i need to
heat the reaction in order to double the equilibrium constant?
1 answer:
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Answer:
0.8
Explanation:
The two spheres have the same potential, V.
Let the radius of the larger sphere be R and the radius of the smaller sphere be r,
=> R = 4r
Let the charge on the smaller sphere be q. Hence, the larger sphere will have charge Q - q.
The potential of the smaller sphere will be:
The potential of the larger sphere will be:
Inputting R = 4r,
Since ,
=> Q - q = 4q
=> 5q = Q
q = 0.2Q
The fraction of the charge Q that rests on the smaller sphere is 0.2
The charge of the larger sphere is:
Q - q = Q - 0.2Q = 0.8Q
∴ The fraction of the total charge Q that rests on the larger sphere is 0.8