In the coal-gasification process, carbon monoxide is converted to carbon dioxide via the following reaction: CO (g) H2O (g) CO2 (g) H2 (

Question

In the coal-gasification process, carbon monoxide is converted to carbon dioxide via the following reaction: CO (g) H2O (g) CO2 (g) H2 (g) In an experiment, 0.35 mol of CO and 0.40 mol of H2O were placed in a 1.00-L reaction vessel. At equilibrium, there were 0.19 mol of CO remaining. Keq at the temperature of the experiment is __________. A) 5.47 B) 1.0 C) 1.78 D) 0.75 E) 0.56

in progress 0
Ben Gia 5 years 2021-08-15T22:57:39+00:00 1 Answers 360 views 1

Answers ( )

  1. Nho
    0
    2021-08-15T22:58:52+00:00
    Reply

    Answer: K_{eq} at  the temperature of the experiment is 0.56.

    Explanation:

    Moles of  CO = 0.35 mole

    Moles of  H_2O = 0.40 mole

    Volume of solution = 1.00 L

    Initial concentration of CO = \frac{0.35mol}{1.00L}=0.35M

    Initial concentration of H_2O = \frac{0.40mol}{1.00L}=0.40M

    Equilibrium concentration of CO = \frac{0.19mol}{1.00L}=0.19M

    The given balanced equilibrium reaction is,

                          CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)

    Initial conc.          0.35 M         0.40 M                  0 M        0M

       
    At eqm. conc.     (0.35-x) M   (0.40-x) M   (x) M      (x) M

    Given:  (0.35-x) = 0.19

    x= 0.16 M

    The expression for equilibrium constant for this reaction will be,

    K_{eq}=\frac{[CO_2]\times [H_2]}{[CO]\times [H_2O]}  

    Now put all the given values in this expression, we get :

    K_{eq}=\frac{0.16\times 0.16}{(0.35-0.16)\times (0.40-0.16)}

    K_{eq}=\frac{0.16\times 0.16}{(0.19)\times (0.24)}=0.56

    Thus K_{eq} at  the temperature of the experiment is 0.56.

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )