We have Kc = 4.2 x 10^-2 (given but missing in the question)
and When the balanced equation for this reaction is:
PCl5(g) ↔ PCl3(g) + Cl2(g)
so, according to the Kc formula:
Kc = the concentration of products / the concentration of the reactants
so, to get the concentration of the reactants in equilibrium, the concentration of the products / the concentration of the reactants should equal the Kc value which is given in the question (missing in your question).
So by substitution in Kc formula:
Kc = [PCl3]*[Cl2] / [PCl5]
4.2 x 10^-2 = 0.18 * 0.25 /[PCl5]
∴[PCl5] = 0.18*0.25 / 4.2x10^-2 = 1.07
So the concentration of the reactants in equilibrim = 1.07
Answer:
Explanation:
mass % of C = 0.27/0.45*100 = 60%
mass % of H = 0.02/0.45*100 = 4.4%
mass % of O = 0.16/0.45*100 = 35.6%
Total = 60%+4.4%+ 35.6% = 100%
Answer:
Explanation:
Discussion
When Pressure increases equilibrium shifts to the side with the smallest number of moles. But which side is that?
N2(g) + 3H2(g) ⇌ 2NH3(g)
The left side has 1 mol of nitrogen (N2) and 3 moles of Hydrogen = 4 mols
on the left side.
The right side has 2 mols of NH3 = 2 mols on the right.
Conclusion: You tell the number of mols by the Balance numbers to the left of each chemical in an equation.
Since the left side N2 + 3H2 = 4 mols, the equilibrium does NOT shift left.
2NH3 is only two mols.
The equilibrium shifts Right
Answer
D
