Question

The water-gas shift reaction is an important source of hydrogen. The value of Kc for the reaction below is is 5.1 at 700 K. CO+Hg) Co(8)+H2(8 ﹀ 1st attempt Part 1 (1point) See Periodic Table Calculate t the equilbrium concentrations of the four gasesif the initial concentration of each of them is 00954 M Calculate the equilibrium concentrations of the four gases if the initial concentration of each of them is 0.0954 M. col Part 2 (1 point) CO21

Answer 1

The reaction given

CO_(g) + H₂O_(g) --------> CO_2(g) + H_2(g)

Kc= 5.1

it is given that initially 0.0954 M of each of them is present

CO_(g)    +           H₂O_(g) -------->            CO_2(g)     +    H_2(g)

0.0954                      0.0954                         0.0954               0.0954

0.0954-x                 0.0954-x                         0.0954+x           0.094+x

Kc expression will be

Keーで에H201

0.0954] 0.0954+ 0.0954- 0.0954-

    5.1 [0.0954-1]*[0.0954-x] = [0.0954+x] * [0.0954+x]

5.1 (0.0954-x)² = (0.0954+x)²

5.1 ( 0.0954² - 2 * 0.0954*x + x² ) = ( 0.0954² + 2 * 0.0954*x + x² )

5.1* 0.0954² -5.1* 2 * 0.0954*x+ 5.1x² = 0.0954² + 2 * 0.0954*x + x²

0.046416 - 0.97308*x + 5.1 x²   = 9.10116*10^-3 +0.1908*x+x²

( 5.1-1)x² -( 0.97308+0.1908)*x +0.046416-9.10116*10^-3 =0

4.1 x² -1.16388x+0.03731484 = 0

on solving the above quadratic equation , x= 0.24703, 0.0368423

Since the initial concentration was 0.0950 , 0.247 cannot be root as that would been negative molar concentration of reactant at equilibrium which is not possible

Hence , x =0.0368423 M

Hence

PART (1)

[CO_(g) ]= 0.0950-x=0.0950-0.0368423 = 0.0581577 M

PART (2)

[CO_2(g) ]= 0.0950+x= 0.0950+0.0368423 = 0.1318423 M

PART (3)

[H₂O_(g)] =0.0950 -x = 0.0950-0.0368423 =0.0581577 M

PART (4)

[H_{2(g) ^]}=0.0950+x = 0.0950+0.0368423 = 0.1318423 M