Question

6. Equimolar amounts of A(g) and B(g) are mixed in a closed container fitted with a moveable piston

(allowing pressure to remain constant). Calculate the ratio of the final volume of the container to the initial
volume of the container when the reaction
A(g) + 4 B(g) - - 3 C(g)
goes to completion at the same temperature as at the onset of the reaction.

Answer 1

The given reaction is:
\[A(g) + 4B(g) \rightarrow 3C(g)\]

From the balanced equation, we can see that for every 1 mole of A that reacts, 4 moles of B are consumed and 3 moles of C are produced.

Since equimolar amounts of A and B are mixed, let's assume we start with 1 mole of each.

At the beginning, we have:
\[A = 1 \, \text{mol}\]
\[B = 1 \, \text{mol}\]
\[C = 0 \, \text{mol}\]

After the reaction goes to completion:
\[A = 0 \, \text{mol}\]
\[B = 0 \, \text{mol}\]
\[C = 3 \, \text{mol}\]

Now, let's consider the ideal gas law equation, which relates the pressure, volume, and amount of gas:
\[PV = nRT\]

Since the pressure and temperature are constant, we can compare the initial and final volumes using the ratios of the moles of gases.

Initial moles of gas (\(n_{\text{initial}}\)) = \(A + B = 1 + 1 = 2\) moles
Final moles of gas (\(n_{\text{final}}\)) = \(C = 3\) moles

The ratio of the final volume (\(V_{\text{final}}\)) to the initial volume (\(V_{\text{initial}}\)) is given by:
\[\frac{V_{\text{final}}}{V_{\text{initial}}} = \frac{n_{\text{initial}}}{n_{\text{final}}}\]

Substitute the values:
\[\frac{V_{\text{final}}}{V_{\text{initial}}} = \frac{2}{3}\]

So, the ratio of the final volume of the container to the initial volume of the container is \(2/3\).