The value of K_c = 4.24 at 800K for the reaction, CO_((g)) + H_2O_((g)) hArrCO_(2(g)) + H_(2(g))Calculate equilibrium concentrations of CO_(2), H_2, CO and H_2O at 800 K, if only CO and H_2O are present initially at concentrations of 0.10M each.

The value of K_c = 4.24 at 800K for the reaction, CO_((g)) + H_2O_((g)

1.	The value of K_c = 4.24 at 800K for the reaction, CO_((g)) + H_2O_((g)) hArrCO_(2(g)) + H_(2(g))Calculate equilibrium concentrations of CO_(2), H_2, CO and H_2O at 800 K, if only CO and H_2O are present initially at concentrations of 0.10M each.
Answer» Solution :`{:("REACTION :", CO_((g))+, H_2O_((g)) hArr , CO_(2(g)) + , H_(2(g))),("INITIAL conc. :",0.1 M,0.1 M, 0 M, 0 M),("Reaction change :",-xM,-X M,+x M,+x M),("Concentration of equilibrium:",(0.1-x)M,(0.1-x)M,x M, x M):}` where , x= amount of CO consume in reaction = amount of `H_2O` x=made of product = amount of `CO_2` = amount `H_2` `K_c=([CO_2][H_2])/([CO][H_2O])=((x)(x))/((0.1-x)(0.1-x))=x^2/(0.1 -x)^2` `therefore x^2/(0.1-x)^2=4.24` `therefore x^2=4.24(0.01-0.2x - x^2)` `therefore x^2=0.0424-0.848x+4.24x^2` `therefore underseta(3.24x^2)-undersetb(0.848x)+undersetc(0.0424)=0` This is quadratic equation in which , a=3.24 , b=-0.848 and c=0.0424 This equation of quadratic , `ax^2+bx+c` `therefore x=(-bpmsqrt(b^2-4ac))/(2a)` `therefore x=(-(-0.848)pmsqrt((0.848)^2-4(3.24)(0.0424)))/(2(3.24))` `therefore x=(+0.848 pm SQRT(0.7191-0.5495))/6.48` `=(0.848pmsqrt(0.1696))/6.48 =(0.848 pm 0.4118)/6.48` `=1.2598/6.48` `0.4362/6.48` =0.1944 OR 0.0673 The value 0.1994 should be neglected because it will give concentration of the reactant which is more than initial conc. So, x=0.0673 M `therefore` Product of equilibrium `[CO_2]=[H_2]`=0.0673 M Reaction of equili. [CO]=`[H_2O]`= (0.1-x)M =(0.1-0.0673)M = 0.0327 M