Question

Suppose a 250. ml. flask is filled with 0.30 mol of IH2 and 1.8 mol of III. The following reaction becomes possible The equilibrium constant &for this reection is 1.87 at the temperature of the flask. Calculate the equilibrium molarity of H, Round your answer to one decimal place The key to solving an equilibcium.compoaition problem is the connection between the equinlbrium KThe sauililacium.constant.aspresion is on the left side of this equation. backwards" to calculate the But If you know K instead, you can use the equation You can use this equation to calculate K from the equilibrium molarities. But if you know equilibrium molerities Firt, set up reaction table: u,1011 [HII] initial Since we're after the equatrium malarity of Hi, we might as wel letitand-u·maw n many of Hi change|xx2x sest suhettuda "e oonresoons in th" last row of th, reaction table for the molarities in the eollhrian

"Calculating equilibrium composition from an equilibrium constant"

Can someone explain how to get the highlighted values?
I'm blanking out right.

Answer 1

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I have provided the description below for the highlighted values as asked in the problem

Molarity = Number of moles of Solute/Volume of Solution (in L)

In our problem, the size of flask is 250 mL

1 L = 1000 mL

so, 250 mL = 0.25L

number of moles of H2 = 0.3 moles

number of moles of HI = 1.8 moles

Molarity of H2 = number of moles of H2/volume of solution in L = 0.3/0.25 = 1.2 M

Molarity of HI = number of moles of HI/volume of solution in L = 1.8/0.25 = 7.2 M

H2 + I2 ------ 2HI

Initial 1.2 0 7.2

Final (1.2-x) x (7.2-2x)