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MISSED THIS? Read Section 16.8 (Pages 701-710); Watch KCV 16.8, IWE 16.9. Consider the reaction CO (g) + H$_2$O (g) $ ightleftharpoons$ CO$_2$ (g) + H$_2$ (g) $K_c$ = 102 at 500 K A reaction mixture initially contains 0.120 M CO and 0.120 M H$_2$O. What will be the equilibrium concentration of CO? Express the concentration in molarity to two significant figures. [CO] = 1.08$ imes10^{-2}$ M Submit Previous Answers Answer Requested An ICE table is helpful to organize concentrations from initial to equilibrium states: [CO] [H$_2$O] [CO$_2$] [H$_2$] Initial 0.120 0.120 0.0 0.0 Change $-x$ $-x$ $+x$ $+x$ Equilibrium 0.120 - $x$ 0.120 - $x$ $x$ $x$ Using the law of mass action yields $K_c = \frac{[CO_2][H_2]}{[CO][H_2O]} = \frac{x^2}{(0.120 - x)^2} = 102$ Solving for $x$ gives [CO] = 0.120 - $x$. Part B What will be the equilibrium concentration of H$_2$O? Express the concentration in molarity to two significant figures. [H$_2$O] = M

          MISSED THIS? Read Section
16.8 (Pages 701-710); Watch KCV 16.8,
IWE 16.9.
Consider the reaction
CO (g) + H$_2$O (g) $
ightleftharpoons$ CO$_2$ (g) + H$_2$ (g)
$K_c$ = 102 at 500 K
A reaction mixture initially contains 0.120 M CO
and 0.120 M H$_2$O.
What will be the equilibrium concentration of CO?
Express the concentration in molarity to two significant figures.
[CO] = 1.08$	imes10^{-2}$ M
Submit Previous Answers
Answer Requested
An ICE table is helpful to organize concentrations from initial to equilibrium states:
[CO]	[H$_2$O]	[CO$_2$]	[H$_2$]
Initial	0.120	0.120	0.0	0.0
Change	$-x$	$-x$	$+x$	$+x$
Equilibrium	0.120 - $x$	0.120 - $x$	$x$	$x$
Using the law of mass action yields
$K_c = \frac{[CO_2][H_2]}{[CO][H_2O]} = \frac{x^2}{(0.120 - x)^2} = 102$
Solving for $x$ gives [CO] = 0.120 - $x$.
Part B
What will be the equilibrium concentration of H$_2$O?
Express the concentration in molarity to two significant figures.
[H$_2$O] = M

MISSED THIS? Read Section
16.8 (Pages 701-710); Watch KCV 16.8,
IWE 16.9.
Consider the reaction
CO (g) + H2O (g) ightleftharpoons CO2 (g) + H2 (g)
Kc = 102 at 500 K
A reaction mixture initially contains 0.120 M CO
and 0.120 M H2O.
What will be the equilibrium concentration of CO?
Express the concentration in molarity to two significant figures.
[CO] = 1.08imes10^-2 M
Submit Previous Answers
Answer Requested
An ICE table is helpful to organize concentrations from initial to equilibrium states:
[CO] [H2O] [CO2] [H2]
Initial 0.120 0.120 0.0 0.0
Change -x -x +x +x
Equilibrium 0.120 - x 0.120 - x x x
Using the law of mass action yields
Kc = ([CO2][H2])/([CO][H2O]) = (x^2)/((0.120 - x)^2) = 102
Solving for x gives [CO] = 0.120 - x.
Part B
What will be the equilibrium concentration of H2O?
Express the concentration in molarity to two significant figures.
[H2O] = M

Added by Cristian T.

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