Question

suppose a 500. ml flask is filled with 0.90 mol of ch₄, 0.30 mol of h₂o and 0.50 mol of co. this reaction becomes possible:
ch₄(g)+h₂o(g)⇌co(g)+3h₂(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 co. you can leave out the m symbol for molarity.

Explanation:

Step1: Calculate initial molarity

Molarity \(c=\frac{n}{V}\), \(V = 500\space mL=0.5\space L\)
For \(CH_{4}\): \(c_{initial}(CH_{4})=\frac{0.90\space mol}{0.5\space L}=1.8\)
For \(H_{2}O\): \(c_{initial}(H_{2}O)=\frac{0.30\space mol}{0.5\space L}=0.6\)
For \(CO\): \(c_{initial}(CO)=\frac{0.50\space mol}{0.5\space L}=1.0\)
For \(H_{2}\): \(c_{initial}(H_{2}) = 0\)

Step2: Determine change in molarity

From the stoichiometry of the reaction \(CH_{4}(g)+H_{2}O(g)
ightleftharpoons CO(g)+3H_{2}(g)\)
If the change in molarity of \(CO\) is \(x\), then:
Change in \(CH_{4}\): \(-x\) (because for every \(1\) mole of \(CO\) produced, \(1\) mole of \(CH_{4}\) is consumed)
Change in \(H_{2}O\): \(-x\) (because for every \(1\) mole of \(CO\) produced, \(1\) mole of \(H_{2}O\) is consumed)
Change in \(H_{2}\): \(+ 3x\) (because for every \(1\) mole of \(CO\) produced, \(3\) moles of \(H_{2}\) are produced)

Step3: Calculate equilibrium molarity

Equilibrium molarity of \(CH_{4}\): \(1.8 - x\)
Equilibrium molarity of \(H_{2}O\): \(0.6 - x\)
Equilibrium molarity of \(CO\): \(1.0+x\)
Equilibrium molarity of \(H_{2}\): \(0 + 3x=3x\)

Answer:

	\(CH_{4}\)	\(H_{2}O\)	\(CO\)	\(H_{2}\)
change	\(-x\)	\(-x\)	\(x\)	\(+3x\)
equilibrium	\(1.8 - x\)	\(0.6 - x\)	\(1.0 + x\)	\(3x\)