Equilibrium
Can someone help me with this question please
For the reaction H2 +I2=2HI
The equilibrium constant has a numerical value of 49.0 at temperature of 150k
If 2.00 moles of hydrogen nad 2.00 moles of iodine are heated in closed vessel at 150k until equilibrium is reached , calculate the composition of the equilibrium mixture in moles
For the reaction H2 +I2=2HI
The equilibrium constant has a numerical value of 49.0 at temperature of 150k
If 2.00 moles of hydrogen nad 2.00 moles of iodine are heated in closed vessel at 150k until equilibrium is reached , calculate the composition of the equilibrium mixture in moles
So initial moles are
H2: 2
I2: 2
2HI:0
at equilibrium, this changes to
H2: 2-x
I2: 2-x
2HI: 2x (=0+2xX)
your expression is:
49 = (2x)^2/(2-x)^2 (you don't need concentrations, since the molar ratio of products: reactants is 1:1.
solve this equation? by square rooting both sides and rearranging. hopefully that will get you the right answer for what x is, and then you can substitute that into the initial expressions for moles at equilibrium.
H2: 2
I2: 2
2HI:0
at equilibrium, this changes to
H2: 2-x
I2: 2-x
2HI: 2x (=0+2xX)
your expression is:
49 = (2x)^2/(2-x)^2 (you don't need concentrations, since the molar ratio of products: reactants is 1:1.
solve this equation? by square rooting both sides and rearranging. hopefully that will get you the right answer for what x is, and then you can substitute that into the initial expressions for moles at equilibrium.
Original post by C_H_B
So initial moles are
H2: 2
I2: 2
2HI:0
at equilibrium, this changes to
H2: 2-x
I2: 2-x
2HI: 2x (=0+2xX)
your expression is:
49 = (2x)^2/(2-x)^2 (you don't need concentrations, since the molar ratio of products: reactants is 1:1.
solve this equation? by square rooting both sides and rearranging. hopefully that will get you the right answer for what x is, and then you can substitute that into the initial expressions for moles at equilibrium.
H2: 2
I2: 2
2HI:0
at equilibrium, this changes to
H2: 2-x
I2: 2-x
2HI: 2x (=0+2xX)
your expression is:
49 = (2x)^2/(2-x)^2 (you don't need concentrations, since the molar ratio of products: reactants is 1:1.
solve this equation? by square rooting both sides and rearranging. hopefully that will get you the right answer for what x is, and then you can substitute that into the initial expressions for moles at equilibrium.
Thank you so much
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