Question

Consider a homogeneous gas phase reaction, \(A + 2B \rightarrow 2C + 3D\), occurring at constant temperature and pressure in a varying volume batch reactor. The vessel is initially charged with 50 moles of A and 150 moles of B. The gases behave ideally. For complete conversion of A, the ratio of the final volume to the initial volume is ______ (rounded off to one decimal place).

Hint:

At constant temperature and pressure, volume is proportional to total moles; find the total moles present after A is completely consumed and compare with the initial total moles.

Answer 1

Correct Answer: 1.5

Step 1: Write the reaction stoichiometry and the initial charge.
Reaction: \(A + 2B \rightarrow 2C + 3D\).
Initial moles: \(n_{A0}=50\), \(n_{B0}=150\), \(n_{C0}=0\), \(n_{D0}=0\).
Total initial moles \(n_0 = 50+150 = 200\) mol.

Step 2: Check which reactant is limiting for complete conversion of A.
Complete conversion of A means all 50 mol of A react. From the stoichiometry, this consumes \(2 \times 50 = 100\) mol of B. Since 150 mol of B is available and only 100 mol is needed, B is in excess and complete conversion of A is achievable, leaving \(150-100=50\) mol of unreacted B.

Step 3: Compute the final mole numbers at complete conversion of A (\(X_A=1\)).
\(n_A = 50 - 50(1) = 0\)
\(n_B = 150 - 2(50)(1) = 150-100 = 50\)
\(n_C = 0 + 2(50)(1) = 100\)
\(n_D = 0 + 3(50)(1) = 150\)
Total final moles \(n_f = 0+50+100+150 = 300\) mol.

Step 4: Relate volume to moles using the ideal gas law.
For an ideal gas, \(PV = nRT\), so \(V = \dfrac{nRT}{P}\). Since the reaction occurs at constant temperature and pressure, \(V\) is directly proportional to \(n\). Therefore:
\[ \frac{V_f}{V_0} = \frac{n_f}{n_0} = \frac{300}{200} = 1.5 \]

Step 5: Cross-check using the fractional volume change (expansion factor) method.
The change in total moles per mole of A reacted is \(\delta = \dfrac{(2+3)-(1+2)}{1} = 2\).
Mole fraction of A initially, \(y_{A0} = \dfrac{50}{200} = 0.25\).
Expansion factor \(\varepsilon_A = y_{A0}\,\delta = 0.25 \times 2 = 0.5\).
For a constant P, T variable-volume batch reactor, \(\dfrac{V}{V_0} = 1+\varepsilon_A X_A\). At \(X_A=1\): \(\dfrac{V}{V_0}=1+0.5(1)=1.5\), which matches Step 4.

\[ \boxed{\dfrac{V_f}{V_0} = 1.5} \]