Question

the equilibrium constant Kp for the reaction H2(g)+Co2(g)—-> H2O+Co(g) is 4.0 at 1660 Initially 0.80 moles of H2 and 0.80 moles of Co2 are injected into 5.0 litre flask what is equilibrium concentration

Answer 1

Answer: Concentration of $H_2O$ = 0.106 M

Concentration of $CO$ = 0.106 M

Concentration of $H_2$ = 0.144 M

Concentration of $H_2O$ = 0.144 M

Explanation:

Initial moles of  $H_2$ = 0.80 mole

Volume of container = 5 L

Initial concentration of $H_2=\frac{moles}{volume}=\frac{0.80moles}{5L}=0.16M$  

Initial moles of  $CO_2$ = 0.80 mole

Volume of container = 5 L

Initial concentration of $CO_2=\frac{moles}{volume}=\frac{0.80moles}{5L}=0.16M$  

Equilibrium concentration of $CO_2=\frac{moles}{volume}=\frac{0.80moles}{5L}=0.16M$

The given balanced equilibrium reaction is,      

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

Initial conc.         0.16         0.16           0         0

At eqm. conc.    (0.16-x)   (0.16-x)       (x)      (x)

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[CO][H_2O]}{[H_2][CO_2]}$

$K_c=\frac{x\times x}{(0.16-x)(0.16-x)}$

we are given :  $K_p=4$

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

Where,

$K_p$ = equilibrium constant in terms of partial pressure = 4

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = $1660K$

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-2=0$

Putting values in above equation, we get:

$4=\frac{x\times x}{(0.16-x)(0.16-x)}$

$x=0.106$

Concentration of $H_2O$ = x = 0.106 M

Concentration of $CO$ = x = 0.106 M

Concentration of $H_2$ = 0.16 - x = 0.16- 0.106 = 0.144 M

Concentration of $H_2O$ = 0.16 - x = 0.16- 0.106 = 0.144 M

Answer 2

the final answer is 0.0533. A little approximation is also valid...