For the water gas shift reaction below, $K_c = 3.491$ at a certain temperature. What are the equilibrium concentrations of all the components of the reaction if $\pu{0.3815 mol}$ of $\ce{CO}$ and $\ce{H2O}$ are initially mixed in a $\pu{250 mL}$ flask?

 

$$\ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)}$$

I was able to calculate the molarity of $\ce{CO}$ and $\ce{H2O}$ and set up an ICE table, but my answer seemed inaccurate. My work so far:

$$3.491 = \frac{[\ce{CO2}][\ce{H2}]}{[\ce{CO}][\ce{H2O}]}$$

$$3.491 = \frac{x^2}{(1.526-x)^2}$$

Because $x$ should be much smaller than $1.5$, it can be ignored when squaring $1.526$. Therefore, $3.491 = x^2/2.32$

$$3.491\cdot 2.32 = x^2$$

The square root of that value is $2.85$, which is impossible because there weren't $\pu{2.85 mol}$ of $\ce{CO}$ or $\ce{H2O}$ to begin with.

For the water gas shift reaction below, $K_c = 3.491$ at a certain temperature. What are the equilibrium concentrations of all the components of the reaction if $\pu{0.3815 mol}$ of $\ce{CO}$ and $\ce{H2O}$ are initially mixed in a $\pu{250 mL}$ flask?

$$\ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)}$$

I was able to calculate the molarity of $\ce{CO}$ and $\ce{H2O}$ and set up an ICE table, but my answer seemed inaccurate. My work so far:

$$3.491 = \frac{[\ce{CO2}][\ce{H2}]}{[\ce{CO}][\ce{H2O}]}$$

$$3.491 = \frac{x^2}{(1.526-x)^2}$$

Because $x$ should be much smaller than $1.5$, it can be ignored when squaring $1.526$. Therefore, $3.491 = x^2/2.32$

$$3.491\cdot 2.32 = x^2$$

The square root of that value is $2.85$, which is impossible because there weren't $\pu{2.85 mol}$ of $\ce{CO}$ or $\ce{H2O}$ to begin with.

For the extremely important water gas shift reaction shown below, the $K_c = 3.491$ at a certain temperature. What are the equilibrium concentrations of all the components of the reaction if $\pu{0.3815 mol}$ of $\ce{CO}$ and $\ce{H2O}$ are initially mixed in a $\pu{250 mL}$ flask?

$$\ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)}$$

I was able to calculate the molarity of $\ce{CO}$ and $\ce{H2O}$ and set up an ICE table, but my answer seemed inaccurate. My work so far:

$$3.491 = \frac{[\ce{CO2}][\ce{H2}]}{[\ce{CO}][\ce{H2O}]}$$

$$3.491 = \frac{x^2}{(1.526-x)^2}$$

Because $x$ should be much smaller than $1.5$, it can be ignored when squaring $1.526$. Therefore, $3.491 = x^2/2.32$

$$3.491\cdot 2.32 = x^2$$

The square root of that value is $2.85$, which is impossible because there weren't $\pu{2.85 mol}$ of $\ce{CO}$ or $\ce{H2O}$ to begin with.

For the water gas shift reaction below, $K_c = 3.491$ at a certain temperature. What are the equilibrium concentrations of all the components of the reaction if $\pu{0.3815 mol}$ of $\ce{CO}$ and $\ce{H2O}$ are initially mixed in a $\pu{250 mL}$ flask?

$$\ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)}$$

I was able to calculate the molarity of $\ce{CO}$ and $\ce{H2O}$ and set up an ICE table, but my answer seemed inaccurate. My work so far:

$$3.491 = \frac{[\ce{CO2}][\ce{H2}]}{[\ce{CO}][\ce{H2O}]}$$

$$3.491 = \frac{x^2}{(1.526-x)^2}$$

Because $x$ should be much smaller than $1.5$, it can be ignored when squaring $1.526$. Therefore, $3.491 = x^2/2.32$

$$3.491\cdot 2.32 = x^2$$

The square root of that value is $2.85$, which is impossible because there weren't $\pu{2.85 mol}$ of $\ce{CO}$ or $\ce{H2O}$ to begin with.

For the extremely important water gas shift reaction shown below, the $K_c = 3.491$ at a certain temperature. What are the equilibrium concentrations of all the components of the reaction if $\pu{0.3815 mol}$ of $\ce{CO}$ and $\ce{H2O}$ are initially mixed in a $\pu{250 mL}$ flask?

$$\ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)}$$

I was able to calculate the molarity of $\ce{CO}$ and $\ce{H2O}$ and set up an ICE table, but my answer seemed inaccurate. My work so far:

$$3.491 = \frac{[\ce{CO2}][\ce{H2}]}{[\ce{CO}][\ce{H2O}]}$$

$$3.491 = x^2/(1.526-x)^2$$

Because $x$ should be much smaller than $1.5$, it can be ignored when squaring $1.526$. Therefore, $3.491 = x^2/2.32$

$$3.491\cdot 2.32 = x^2$$

The square root of that value is $2.85$, which is impossible because there weren't $\pu{2.85 mol}$ of $\ce{CO}$ or $\ce{H2O}$ to begin with.

For the extremely important water gas shift reaction shown below, the $K_c = 3.491$ at a certain temperature. What are the equilibrium concentrations of all the components of the reaction if $\pu{0.3815 mol}$ of $\ce{CO}$ and $\ce{H2O}$ are initially mixed in a $\pu{250 mL}$ flask?

$$\ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)}$$

I was able to calculate the molarity of $\ce{CO}$ and $\ce{H2O}$ and set up an ICE table, but my answer seemed inaccurate. My work so far:

$$3.491 = \frac{[\ce{CO2}][\ce{H2}]}{[\ce{CO}][\ce{H2O}]}$$

$$3.491 = \frac{x^2}{(1.526-x)^2}$$

Because $x$ should be much smaller than $1.5$, it can be ignored when squaring $1.526$. Therefore, $3.491 = x^2/2.32$

$$3.491\cdot 2.32 = x^2$$

The square root of that value is $2.85$, which is impossible because there weren't $\pu{2.85 mol}$ of $\ce{CO}$ or $\ce{H2O}$ to begin with.