1) According to Le Châtelier\'s principle, if the volume of the vessel containin
ID: 878536 • Letter: 1
Question
1)
According to Le Châtelier's principle, if the volume of the vessel containing the equilibrium system shown below is decreased, there will be an increase in the concentration of ________ and a decrease in the concentration of ________.
H2(g) 2 H(g)
Express your answers as chemical formulas separated by a comma.
*please explain.
2)
The equilibrium constant, K c, is equal to 1.4 at 1200° K for the reaction:
CO2(g) + H2(g) CO(g) + H2O(g)
If 0.50 moles of CO2 and 1.0 moles of H2 are introduced into a 1.0-L flask, what will be the concentration of CO when equilibrium is reached?
* I know this is an ICE table equation, but I seem to lack knowledge of the quadratic equation part, this is where things go south. Please make sure to explain this in detail
Explanation / Answer
1)
According to Le Châtelier's principle, if the volume of the vessel containing the equilibrium system shown below is decreased, there will be an increase in the concentration of H2 and a decrease in the concentration of H(g).
H2(g) 2 H(g)
Answer: If the volume of system is decreases then concentration of the system will increase (moles / volume), so in order to decrease the concentration the system will try to shift in the direction where number of moles are less. As shown in the equation the number of moles of gaseous species is less on reactant side hence the system will move backward and there will be a decrease in concentration of H(g) and increase in concentration of H2(g)
2)
The equilibrium constant, K c, is equal to 1.4 at 1200° K for the reaction:
CO2(g) + H2(g) CO(g) + H2O(g)
Initial 0.5 1 0 0
change -x -x +x +x
Equilibrium 0.5-x 1-x x x
Kc = [CO][H2O]/[CO2][H2] = 1.4
1.4 = (x)2 / (0.5-x)(1-x)
1.4= (x)2 / 0.5 -0.5x -x +x2
1.4x2 + 0.7 - 2.1x = x2
0.4x2 + 0.7 - 2.1x = 0
On solving
x = 0.36 = concentration of CO when equilibrium is reached.
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