when 2.75 mol HI(g) placed in 1L container and allowed to dissociate into
2HI(g) <==> H2(g)+I2(g)
Final H2 concin .275M. What is Ka for rxn
The answer is .0156. But I don't know why
Thanks
..........2HI ==> H2 + I2
initial...2.75.....0....0
change...-2x........x....x
equil....2.75-2x....x.....x
The problem tells you H2 (which I've let = x above) = 0.275; therefore, I2 must be 0.275 and HI must be 2.75-(2*0.275) = ?
Now just substitute into Ka expression and solve. I get 0.0156 also.
could you do it in moles instead of molars
and in the K eqn is the M for HI squared
2.75 moles in 1L = 2.75M so the solution I posted is both moles and molar.
Yes, (HI) is squared.
Ka = (H2)(I2)/(HI)^2
Ka = (0.275)(0.275)/(2.20)^2
Ka = 0.0156
To find the equilibrium constant (Ka) for the given reaction, we need to use the information provided about the initial and final concentrations. First, let's set up an ICE (initial, change, equilibrium) table to keep track of the concentrations during the reaction.
Initial:
HI(g): 2.75 mol
H2(g): 0 mol
I2(g): 0 mol
Change:
HI(g): -2x
H2(g): +x
I2(g): +x
Equilibrium:
HI(g): 2.75 - 2x
H2(g): x
I2(g): x
From the problem, we also know that the final concentration of H2(g) is 0.275 M. Plugging this value into the equilibrium row, we get:
x = 0.275
Now we can use this value of x to calculate the equilibrium concentration of HI(g), which is 2.75 - 2x:
HI(g) = 2.75 - 2(0.275)
HI(g) = 2.75 - 0.55
HI(g) = 2.2
Now we can substitute the equilibrium concentrations into the expression for Ka:
Ka = ([H2][I2])/[HI]^2
Plugging in the values:
Ka = (0.275 * 0.275)/(2.2^2)
Ka = 0.075625/4.84
Ka = 0.01557 (approximately)
Therefore, the equilibrium constant (Ka) for the given reaction is approximately 0.0156.