Question

suppose a 250. ml flask is filled with 1.0 mol of co, 2.0 mol of co₂ and 1.3 mol of h₂. this reaction becomes possible:
co(g) + h₂o(g) ⇌ co₂(g) + h₂(g)
complete the table below, so that it lists the initial molarity of each compound, the change in molarity of each compound due to the reaction, and the equilibrium molarity of each compound after the reaction has come to equilibrium.
use x to stand for the unknown change in the molarity of h₂o. you can leave out the m symbol for molarity.

Explanation:

Step1: Calculate initial molarity

The formula for molarity \(M=\frac{n}{V}\), where \(n\) is the number of moles and \(V = 250\space mL=0.25\space L\)

• For \(CO\): \(M_{CO}=\frac{1.0\space mol}{0.25\space L} = 4.0\)
• For \(H_{2}O\): Initially \(n = 0\), so \(M_{H_{2}O}=0\)
• For \(CO_{2}\): \(M_{CO_{2}}=\frac{2.0\space mol}{0.25\space L}=8.0\)
• For \(H_{2}\): \(M_{H_{2}}=\frac{1.3\space mol}{0.25\space L}=5.2\)

Step2: Determine change in molarity

Based on the stoichiometry of the reaction \(CO(g)+H_{2}O(g)
ightleftharpoons CO_{2}(g)+H_{2}(g)\) (mole ratio \(1:1:1:1\))

• Change for \(CO\): \(-x\) (since it reacts with \(H_{2}O\))
• Change for \(H_{2}O\): \(+x\) (given)
• Change for \(CO_{2}\): \(+x\)
• Change for \(H_{2}\): \(+x\)

Step3: Calculate equilibrium molarity

• Equilibrium for \(CO\): \(4.0 - x\)
• Equilibrium for \(H_{2}O\): \(0 + x=x\)
• Equilibrium for \(CO_{2}\): \(8.0+x\)
• Equilibrium for \(H_{2}\): \(5.2 + x\)

Answer:

	\(CO\)	\(H_{2}O\)	\(CO_{2}\)	\(H_{2}\)
Change	\(-x\)	\(+x\)	\(+x\)	\(+x\)
Equilibrium	\(4.0 - x\)	\(x\)	\(8.0 + x\)	\(5.2 + x\)