2.00 mol of H2(g) and 1.00 mol of I2(g) are placed in a 1.00 L container, and they react to form HI(g). At equilibrium, it is found that 1.80 moles of HI(g) are present in the container. Calculate K for the reaction: H2(g) + I2(g) ⇄ 2 HI(g)

The value of equilibrium constant is 29.45.

Explanation:

Moles of hydrogen gas = 2.00 mol

Concentration of  hydrogen gas =

Moles of iodine gas = 1.00 mol

Concentration of  iodine gas =

initially

2.00 M   1.00 M    1.00 M

At equilibrium:

(2.00-x/2)    (1.00-x/2)      x

Moles of  HI at equilibrium = 1.80 M

Concentration of HI at equilibrium =

The expression of an equilibrium constant is given by ;

Putting x equal to 1.80 M.

The value of equilibrium constant is 29.45.

*tilts head* sounds more like a statement

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 :