# If 0.01 mol of H_2 and 0.01 mol of CO_2 are mixed in a 1 litre container at 750°C, what are the concentrations of all substances (in mol/L) present at equilibrium?

## given the reaction is: $C {O}_{2} + {H}_{2} \leftrightarrow C O + {H}_{2} O$ and the equilibrium constant, K = 0.771 Also, What is the partial pressure of $C O$ in the container at equilibrium?

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
Oct 20, 2017

concentration of CO and ${H}_{2} O$= $4 , 67 \times {10}^{-} 3 \frac{m o l}{L}$
concentration of ${H}_{2}$ and $C {O}_{2}$=$5 , 33 \times {10}^{-} 3 \frac{m o l}{L}$

#### Explanation:

let's call x the mol of ${H}_{2}$ which react (and remain at the equilibrium 0,01-x), it will be x also X the mol of $C {O}_{2}$ that react (and remain at the equilibrium 0,01-x). they will form x mol of CO and x mol of water
we can write: ${K}_{c} = \frac{X \times X}{\left(0 , 01 - X\right) \left(0 , 01 - X\right)} = 0 , 771$
${X}^{2} = 0 , 771 {\left(0 , 01 - X\right)}^{2}$
by solving the second grade equation you obtain $X = 4 , 67 \times {10}^{-} 3$ that are the concentration in mol/L of both CO and ${H}_{2} O$ while the concentration of ${H}_{2}$ and $C {O}_{2}$ wil be $\left(1 - 0 , 467\right) {10}^{2} = 0 , 533 \times {10}^{- 2} = 5 , 33 \times {10}^{-} 3 \frac{m o l}{L}$

• 12 minutes ago
• 15 minutes ago
• 19 minutes ago
• 22 minutes ago
• 59 seconds ago
• A minute ago
• A minute ago
• 5 minutes ago
• 7 minutes ago
• 7 minutes ago
• 12 minutes ago
• 15 minutes ago
• 19 minutes ago
• 22 minutes ago