Answer:
0.44 moles
Explanation:
Given that :
A mixture of water and graphite is heated to 600 K in a 1 L container. When the system comes to equilibrium it contains 0.17 mol of H2, 0.17 mol of CO, 0.74 mol of H2O, and some graphite.
The equilibrium constant [tex]K_c= \dfrac{[CO][H_2]}{[H_2O]}[/tex]
The equilibrium constant [tex]K_c= \dfrac{(0.17 )(0.17)}{0.74}[/tex]
The equilibrium constant [tex]K_c= 0.03905[/tex]
Some O2 is added to the system and a spark is applied so that the H2 reacts completely with the O2.
The equation for the reaction is :
[tex]H_2 + \dfrac{1}{2}O_2 \to H_2O \\ \\ 0.17 \ \ \ \ \ \ \ \ \ \to0.17[/tex]
Total mole of water now = 0.74+0.17
Total mole of water now = 0.91 moles
Again:
[tex]K_c= \dfrac{[CO][H_2]}{[H_2O]}[/tex]
[tex]0.03905 = \dfrac{[0.17+x][x]}{[0.91 -x]}[/tex]
0.03905(0.91 -x) = (0.17 +x)(x)
0.0355355 - 0.03905x = 0.17x + x²
0.0355355 +0.13095 x -x²
x² - 0.13095 x - 0.0355355 = 0
By using quadratic formula
x = 0.265 or x = -0.134
Going by the value with the positive integer; x = 0.265 moles
Total moles of CO in the flask when the system returns to equilibrium is :
= 0.17 + x
= 0.17 + 0.265
= 0.435 moles
=0.44 moles (to two significant figures)
