Suppose a 250. mL flask is filled with 0.50 mol of H, and 0.90 mol of Hl. This reaction becomes possible: H, (g)+I,(g) – 2HI(g) Complete the table below, so that it lists the initial molarity of each compound, the change in molarity of each compound due to the reaction, and the equilibrium molarity of each compound after the reaction has come to equilibrium. Use x to stand for the unknown change in the molarity of H,. You can leave out the M symbol for molarity. H, HI initial change equilibrium

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Suppose a 250. mL flask is filled with 0.50 mol of H, and 0.90 mol of HI. This reaction becomes possible:
H, (g) +1,(g) = 2HI (g)
Complete the table below, so that it lists the initial molarity of each compound, the change in molarity of each compound due to the reaction, and the
equilibrium molarity of each compound after the reaction has come to equilibrium.
Use x to stand for the unknown change in the molarity of H,. You can leave out the M symbol for molarity.
H2
HI
initial
?
change
equilibrium
Transcribed Image Text:Suppose a 250. mL flask is filled with 0.50 mol of H, and 0.90 mol of HI. This reaction becomes possible: H, (g) +1,(g) = 2HI (g) Complete the table below, so that it lists the initial molarity of each compound, the change in molarity of each compound due to the reaction, and the equilibrium molarity of each compound after the reaction has come to equilibrium. Use x to stand for the unknown change in the molarity of H,. You can leave out the M symbol for molarity. H2 HI initial ? change equilibrium
Expert Solution
Step 1

Given: Volume of flask = 250 mL = 0.250 L                                     (Since 1 L = 1000 mL)

Initial moles of H2 = 0.50 mol.

And initial moles of HI = 0.90 mol.

The equilibrium reaction given is,

=> H2 (g)+ I2 (g)  2 HI(g)

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