Given the following information, what is the concentration of h2o(g) at equilibrium? [h2s](eq) = 0.671 m [o2](eq) = 0.587 m kc = 1.35 2h2s(g) + o2(g) ⇌ 2s(s) + 2h2o(g)

The equilibrium concentration of water is 0.597 M

Explanation:

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as

For a general chemical reaction:

The expression for is written as:

The concentration of pure solids and pure liquids are taken as 1 in the expression.

For the given chemical reaction:

The expression of for above equation is:

We are given:

Putting values in above expression, we get:

Hence, the equilibrium concentration of water is 0.597 M

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 :