Question

The value of kc = 4.24 at 800k for the reaction co(g) + h2o(g) co2(g) + h2(g) calculate the equilibrium concentration of co2 ,h2o ,co , h2 at 800k ,if only co and h2o are present initially at concentration of 0.10m each.

Answer 1

Hey there,

CO(g)+ H2O(g) = CO2(g)+H2(g)

= [2 ][2] /[][2]  
= × /(0.1 − )(0.1 − )
= 4.24 

² = 4.24 × (0.1 − )²
² = 4.24 × (0.01 − 0.2 + ² ) 
² = 0.0424 − 0.848 + 4.24²
3.24²+ 0.0424 − 0.848 = 0
² = 0.194 
= 0.067

Hope this helps!

Answer 2

For the reaction, CO (g) + H2O (g) ⇔ CO2 (g) + H2 (g) Initial concentration: 0.1M 0.1M 0 0 Let x mole per litre of each of the productbe formed. At equilibrium: 0.1 - x M 0.1 - x M x M x M where x is the amount of CO2and H2 at equilibrium.. Hence, equilibrium constant can be written as, Kc= x2/(0.1-x)2 = 4.24 x2 = 4.24(0.01 + x2-0.2x) x2= 0.0424 + 4.24x2-0.848x 3.24x2 – 0.848x + 0.0424 = 0 a = 3.24, b = – 0.848, c = 0.0424 for quadratic equation ax2 + bx + c =0, Thus solving we get two values of x x1= 0.067 x2= 0.194 Neglecting x2= 0.194 becuase x could not be more than initial concentration. Hence the equilibrium concentrations are, [CO2] = [H2] = x = 0.067M [CO] = [H2O] = 0.1 – 0.067 = 0.033 M