Question

a mixture of gases containing 0.20 mol of so₂ and 0.20 mol of o₂ in a 4.0 l flask reacts to form so₃ according to the equation below. 2so₂(g) + o₂(g) → 2so₃(g) if the temperature is 25°c, what is the pressure in the flask after reaction is complete? a 0.3(0.082)(298)/4 atm b 0.2(0.082)(298)/4 atm c 0.4(0.082)(298)/4 atm d 0.3(0.082)(298)/4 atm

Explanation:

Step1: Determine the limiting reactant

From the balanced equation \(2SO_{2}(g)+O_{2}(g)\to2SO_{3}(g)\), the mole ratio of \(SO_{2}\) to \(O_{2}\) is \(2:1\).
We have \(n(SO_{2}) = 0.20\space mol\) and \(n(O_{2})=0.20\space mol\).
For \(SO_{2}\) to react completely, it needs \(n(O_{2})=\frac{0.20}{2}=0.10\space mol\).
Since we have \(0.20\space mol\) of \(O_{2}\), \(SO_{2}\) is the limiting reactant.

Step2: Calculate the moles of products and remaining reactant

After reaction, moles of \(SO_{3}\) formed: Using the mole ratio \(2SO_{2}\to2SO_{3}\), \(n(SO_{3}) = 0.20\space mol\)
Moles of \(O_{2}\) remaining: \(n(O_{2})_{remaining}=0.20 - 0.10=0.10\space mol\)
Total moles of gas \(n = n(SO_{3})+n(O_{2})_{remaining}=0.20 + 0.10=0.30\space mol\)

Step3: Use the ideal gas law \(PV = nRT\)

We know \(V = 4.0\space L\), \(n = 0.30\space mol\), \(R=0.0821\space L\cdot atm/(mol\cdot K)\), \(T=(25 + 273)K=298\space K\)
From \(P=\frac{nRT}{V}\), substituting the values \(P=\frac{0.3\times0.0821\times298}{4}\space atm\)

Answer:

A. \(\frac{0.3(0.0821)(298)}{4}\space atm\)