2) show the calculation of kc for the following reaction if an initial reaction mixture of 0.800 mole of co and 2.40 mole of h2 in a 8.00 liter container forms an equilibrium mixture containing 0.309 mole of h2o and corresponding amounts of co, h2, and ch4.

Kc = 3.90

Explanation:

CO reacts with to form and . balanced reaction is:

No. of moles of CO = 0.800 mol

No. of moles of = 2.40 mol

Volume = 8.00 L

Concentration =

Concentration of CO =

Concentration of =

Initial            0.100      0.300             0   0

equi.            0.100 -x    0.300 - 3x     x    x

It is given that,

at equilibrium = 0.309/8.00 = 0.0386 M

So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M

At equilibrium = 0.300 - 0.0386 × 3 = 0.184 M

At equilibrium = 0.0386 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 :