Question

4. For the reversible reaction 90g) + H40 (g) 500, (g) + H2 (g) a. Write the expression for the equilibrium constant or the equilibrium constant of this reaction. b. Find the equilibrium constant for this expression. The equino are: |CO| 2.00 mol/L; [H.00.241 mol/L: [CO] 3.52 mol/L; [H.] 4.30 mol/L pression. The equilibrium concentrations at 861°C c. If a 10.0 L vessel has 2.50 mol of CO and H-O and 5.00 mol of CO, and H. at 861°C, which way will the reaction shift? (Hint: Calculate Q) d. Using the values from parte, create an ICE table for this system, but do not solve it.

Answer 1

Consider given reaction, CO (g) + H₂O (g) $\rightleftharpoons$ CO ₂ (g) + H ₂ (g)

1) For above reaction, K c = [CO ₂] [H ₂] / [CO] [H₂O]

For above reaction, K p = (P _{CO 2} ) (P _{H 2} ) / ( P _CO) ( P _H2O)  

2) We have , [CO] = 2.00 M , [H ₂]= 4.30 M , [CO ₂] = 3.52 M & [H₂O] = 0.241 M

Putting above given values into equilibrium constant expression, we get

K c = ( 3.52 M ) ( 4.30 M ) / ( 2.00 M) ( 0.241 M) = 31.4

ANSWER : K c = 31.4

3) We have, [CO] = No. of moles / Volume in L

$\therefore$ [CO] = [H₂O] = 2.50 mol / 10.0 L = 0.250 M

Similarly, [CO ₂] [H ₂] = 5.00 mol / 10.0 L = 0.500 M

For above reaction, Q c = [CO ₂] [H ₂] / [CO] [H₂O]

$\therefore$ Q c = ( 0.500 ) ( 0.500) / ( 0.250 ) ( 0.250 ) = 4

Q c < K c means concentrations of products are less than equilibrium concentrations. Hence, to achieve equilibrium reaction will proceed in forward direction to increase concentration of products.

ANSWER : Reaction will shift to the right side. ( product side or forward reaction )

4) Let's use ICE table

Concentration (M)	CO (g) + H2O (g) $\rightleftharpoons$   CO 2 (g) + H 2 (g)
Initial	0.250	0.250	0.500	0.500
Change	-X	-X	+X	+X
Equilibrium	0.250 -X	0.250 -X	0.500+ X	0.500 +X