# Fill The ICE table, calculate the concentration of each gas, write Kc expression..?

Jul 30, 2018
• K_c = (["CO"(g)] * ["H"_2"O" (g)])/(["CO"_2(g)] * ["H"_2(g)]
• c("CO") = c("H"_2"O")~~ 1.356 color(white)(l) "mol"
• c("CO"_2) = c("H"_2)~~ 2.644 color(white)(l) "mol"

#### Explanation:

The equilibrium constant is the ratio of the product of the concentration of products ($\text{CO} \left(g\right)$ and $\text{H"_2"O} \left(g\right)$) to that of the reactants (${\text{CO}}_{2} \left(g\right)$ and ${\text{H}}_{2} \left(g\right)$).

$\textcolor{w h i t e}{l l}$ ${\text{CO"(g) + "H"_2"O"(g) rightleftharpoons "CO"_2(g) + "H}}_{2} \left(g\right)$
I color(white)(l) 2.00 color(white)(l(g) +) 2.00 color(white)(l_2(g) rightleftharpoons) 2.00 color(white)(g")" +)2.00
C -x color(white)(l(g) +) -x color(white)(l(g) rightleftharpoons ) +x color(white)(g")" +) +x
E $2.00 - x \textcolor{w h i t e}{l l l} 2.00 - x \textcolor{w h i t e}{r i g h t \le f t h a r p \infty n s} 2.00 + x \textcolor{w h i t e}{l} 2.00 + x$

$\frac{{\left(2.00 + x\right)}^{2}}{{\left(2.00 - x\right)}^{2}} = {K}_{c} = 3.80$

$x \approx 0.644 \textcolor{w h i t e}{l} \text{M}$

To calculate equilibrium concentrations, substitute $x$ into expressions in the last row of the ICE table:

• c("CO") = 2 - x ~~ 1.356 color(white)(l) "M"
• c("H"_2"O") = 2 - x~~ 1.356 color(white)(l) "M"

• c("CO"_2) = 2 + x ~~ 2.644 color(white)(l) "M"

• c("H"_2) = 2 + x ~~ 2.644 color(white)(l) "M"