Question

In: Chemistry

The equilibrium constant, K, for the reaction below is 57.0 at 700K Hydrogen gas plus iodine...

The equilibrium constant, K, for the reaction below is 57.0 at 700K

Hydrogen gas plus iodine gas yields hydrogen iodide gas.

Write the balanced equation for the reaction. If 1.00 mol of hydrogen gas is allowed to react with 1.00 moles of iodine gas in a 10.0 Liter container, what are the concentrations of each gas at equilibrium?

Expert Solution

Given H₂(g) + I₂(g) 2HI ; K = 57.0 at 700 K

Molarity of H₂ is M =

Molarity of I₂ is M' =

H₂(g) + I₂(g) 2HI

initial mole 1.00 1.00 0

Eqb mole 1.00 -a 1.00 -a 2a Here 2a = number of moles of HI at Equilibrium

Eqb conc (in mol/L)

Equilibrium constant ,

$57.0=\frac{\left \left ( \frac{2a}{10} \right )^{2}}{\left \left ( \frac{1-a}{10} \right )\left ( \frac{1-a}{10}})$

On solving we get a = 0.79 mol/L

Equilibrium concentration of HI = 2a = 2x0.79 = 1.58 mol/L

Equilibrium concentration of H₂ =

Equilibrium concentration of I₂ =

queen_honey_blossom answered 4 months ago

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