This reaction was at equilibrium when 0.2 atm of iodine gas was pumped into the container, what happened to the equilibrium and the partial pressures of the gases

Q was < K. Partial pressure of hydrogen decreased, iodine increased

Explanation:

After iodine was added the Q was [Select] K so the reaction shifted toward the Products [Select] ,The partial pressure of hydrogen [Select], Iodine [Select] |,and hydrogen iodide Decreased

Based on the equilibrium:

H2(g) + I2(g) ⇄ 2HI(g)

K of equilibrium is:

K = [HI]² / [H2] [I2]

Where [] are concentrations at equilibrium

And Q is:

Q = [HI]² / [H2] [I2]

Where [] are actual concentrations of the reactants.

When the reaction is in equilibrium, K=Q.

But as [I2] is increased, Q decreases and Q was < K

The only concentration that increases is [I2], doing partial pressure of hydrogen decreased, iodine increased

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :