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 = 2/3

the equilibrium constant k for this equilibrium is 2/3. predict the number of r and p molecules in this sample when the interconversion reaches equilibrium.

number of r molecules:

number of p molecules:

Explanation:

Step1: Count the initial number of molecules

From the diagram, initially, there are \(6\) \(P\) molecules and \(4\) \(R\) molecules. The total number of molecules \(n = 6 + 4=10\).

Step2: Set up the equilibrium - constant expression

The equilibrium - constant expression for the reaction \(R(aq)
ightleftharpoons P(aq)\) is \(K=\frac{[P]}{[R]}\). Since the volume is the same for both \(R\) and \(P\) (because it's an aqueous solution), we can use the number of molecules instead of molarity. Let \(x\) be the number of \(R\) molecules that react to form \(P\) (or vice - versa). At equilibrium, the number of \(R\) molecules is \(4 + x\) and the number of \(P\) molecules is \(6 - x\) (assuming \(R\) is being formed from \(P\)). And \(K = \frac{2}{3}=\frac{6 - x}{4 + x}\).

Step3: Solve the equation for \(x\)

Cross - multiply: \(2(4 + x)=3(6 - x)\).
Expand: \(8 + 2x=18-3x\).
Add \(3x\) to both sides: \(8 + 2x+3x=18-3x + 3x\), so \(8 + 5x=18\).
Subtract \(8\) from both sides: \(5x=18 - 8=10\).
Divide by \(5\): \(x = 2\).

Step4: Calculate the number of molecules at equilibrium

The number of \(R\) molecules at equilibrium: \(n_R=4 + 2=6\).
The number of \(P\) molecules at equilibrium: \(n_P=6 - 2 = 4\).

Answer:

number of \(R\) molecules: \(6\)
number of \(P\) molecules: \(4\)