1)The Kc for the following reaction is 6.50 at 600 K. CO(g) + H2O(g) CO2(g) + H2
ID: 1004941 • Letter: 1
Question
1)The Kc for the following reaction is 6.50 at 600 K.
CO(g) + H2O(g) CO2(g) + H2(g)
If 2.00 mole of CO and 2.00 mole of H2O are placed in a 1.00 L flask at 600 K, calculate the CO concentration after equilibrium is reached.
2)Consider the following equilibrium:
PCl5 (g) PCl3 (g) + Cl2 (g)
5.0 moles of PCl5 are placed in a 10.0 L flask at 200°C and allowed to come to equilibrium. Analysis
shows that 1.0 mole of Cl2 is present in the equilibrium mixture. How many moles of PCl5 are
present at equilibrium.
Hi, can someone please help me with these problems and explain them as simple as possible please? Thank you!
Explanation / Answer
Solution:- (1) The given balanced rquation is...
CO(g) + H2O(g) <----------------> CO2(g) + H2(g)
Initially we have 2.00 mole of CO and 2.00 mole of H2O and th evolume of the flask is 1.00 L. So, the concentrations for each of the reactants would be 2.00M.
let's say the change in concentration is X. So, the ice table would be...I stands for initial concentration, C stands for change in concentration and E stands for equilibrium concentrations..
CO(g) + H2O(g) <----------------> CO2(g) + H2(g)
I 2.00 2.00 0 0
C -X -X +X +X
E 2.00-X 2.00-X X X
Kc = [CO2][H2]/([CO] [H2O])
Let's plug in the values in it...
6.50 = [X]2/[2.00-X]2
taking square root to both sides...
2.55 = X/(2.00-X)
doing cross multiply...
2.55(2.00-X) = X
5.10 - 2.55X = X
on rearrangement...
5.10 = X + 2.55X
5.10 = 3.55X
X = 5.10/3.55 = 1.44
So, the equilibrium concentration of CO is = 2.00 - 1.44 = 0.56M (answer)
(2) balanced equation is given as..
PCl5(g) <-------------> PCl3(g) + Cl2(g)
Initial moles of PCl5 are 5.0 and the volume of the flask is 10.0 L. So, the initial concentration of PCl5 is 5.0mol/10.0L = 0.50 M
Let's say the change is X and make the ice table..
PCl5(g) <-------------> PCl3(g) + Cl2(g)
I 0.50 0 0
C -X +X +X
E 0.50-X X X
from ice table, at equilibrium the concentration of Cl2 is X and from the information given, at equilibrium 1.0 mole of Cl2 was found. so, equilibrium concentration of Cl2 would be = 1.0 mol/10.0L = 0.10 M
It means, X = 0.10M
Also, from the ice table, at equilibrium, concentration of PCl5 = 0.50 - X
Let's plug in the value of X here...
[PCl5] = 0.50M - 0.10M = 0.40M or 0.40mole/L
we are asked to calculate the equilibrium moles of PCl5. So, let's multiply the concentration by volume.
Hence, moles of PCl5 at equilibrium = (0.40 mole/L) x 10.0 L = 4.0 mole. (Answer)
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