QUESTION IMAGE
Question
a tiny sample of an aqueous solution of two substances r and p is sketched below, as if it was under an imaginary microscope so powerful that individual molecules could be seen. (the water molecules are not shown.)
r and p can interconvert. that is, r can turn into p, and p can turn back into r:
r(aq) ⇌ p(aq) k = 4
the equilibrium constant k for this equilibrium is 4. predict the number of r and p molecules in this sample when the interconversion reaches equilibrium.
number of r molecules:
number of p molecules:
Step1: Count the initial number of molecules
From the sketch, initially, there are \(n_{R,initial}=10\) (R molecules) and \(n_{P,initial} = 1\) (P molecule). The total number of molecules \(N=n_{R}+n_{P}=10 + 1=11\) (this number remains constant during the inter - conversion since the reaction is \(R(aq)
ightleftharpoons P(aq)\) and there is no addition or removal of matter from the system).
Step2: Set up the equilibrium expression
The equilibrium constant \(K=\frac{[P]}{[R]}\) (for the reaction \(R(aq)
ightleftharpoons P(aq)\)). Since the volume is the same for both \(R\) and \(P\) (because it is an aqueous solution), we can use the ratio of the number of molecules. Let the number of \(R\) molecules at equilibrium be \(x\), then the number of \(P\) molecules at equilibrium is \(11 - x\).
We know that \(K = 4=\frac{11 - x}{x}\)
Step3: Solve the equation for \(x\)
Cross - multiply the equation \(4=\frac{11 - x}{x}\):
Add \(x\) to both sides of the equation:
(This is wrong, we made a mistake in initial count. Let's recount the initial molecules. From the sketch, initially \(n_{R,initial}=9\) and \(n_{P,initial}=1\), so \(N = 9 + 1=10\))
Let the number of \(R\) molecules at equilibrium be \(x\), then the number of \(P\) molecules at equilibrium is \(10 - x\).
Since \(K = 4=\frac{10 - x}{x}\)
Cross - multiply: \(4x=10 - x\)
Add \(x\) to both sides: \(4x+x=10\)
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number of \(R\) molecules: \(2\)
number of \(P\) molecules: \(8\)