Question

At a certain temperature, Kc equals 1.4 × 102 for the reaction: 2 CO(g) + O2(g) ⇌ 2 CO2(g). If a 3.50-L flask contains 0.400 mol of CO2 and 0.100 mol of O2 at equilibrium, how many moles of CO are also present in the flask?

Answer 1

Answer:

$\boxed{\text{0.200 mol}}$

Explanation:

The balanced equation is

2CO + O₂ ⇌ 2CO₂

Step 1. Calculate the known concentrations.

$\text{[O _{2} ]} = \dfrac{\text{0.100 mol}}{\text{3.50 L}} = \text{0.02857 mol/L}\\\\\text{[CO _{2} ]} = \dfrac{\text{0.400 mol}}{\text{3.50 L}} = \text{0.1143 mol/L}$

Step 2. Calculate the concentration of CO

$K_{\text{eq}} = \dfrac{\text{[CO _{2} ] ^{2} }}{\text{[CO] ^{2} [O _{2} ]}} = 1.4 \times10^{2}$

$\begin{array}{rcl}\\\\\dfrac{0.1143^{2}}{\text{[CO] ^{2} } \times 0.02857} & = & 1.4 \times10^{2}\\\\0.01306 & = & 4.000\text{[CO] ^{2} }\\\\\text{[CO] ^{2} } & = &\dfrac{0.01306}{4.000}\\\\\text{[CO] ^{2} } & = & 0.003265\\\text{[CO]} & = & \mathbf{0.05714}\\\end{array}$

3. Calculate the moles of CO

n = 3.50 L × 0.057 14 mol·L⁻¹ = 0.200 mol

$\text{At equilibrium, there are }\boxed{\textbf{0.200 mol}} \text{ of CO in the flask}$

Answer 2

Kc is the equilibrium constant and depicts the reactants or the product concentration. At equilibrium, the moles of the carbon monoxide is 0.200 moles.

What are moles?

Moles is the ratio of the mass of the substance to that of the molar mass of the substance.

The balanced chemical reaction is given as,

$\rm 2CO + O_{2} \rightleftharpoons 2CO_{2}$

The known concentration of oxygen is calculated as:

$\begin{aligned}\rm [O_{2}] &= \dfrac{0.100\;\rm mol}{3.50\;\rm L}\\\\&= 0.0285\;\rm mol/L\end{aligned}$

The known concentration of carbon dioxide is calculated as:

$\begin{aligned}\rm [CO_{2}] &= \dfrac{0.400\;\rm mol}{3.50\;\rm L}\\\\&= 0.1143\;\rm mol/L\end{aligned}$

The concentration of the carbon monoxide is calculated as:

$\begin{aligned}\rm k_{eq} = \rm \dfrac{[CO_{2}]^{2}}{[CO]^{2}[O_{2}]} &= 1.4 \times 10^{2}\\\\\rm \dfrac{(0.1143)^{2}}{[CO]^{2}\times 0.0285} &= 1.4 \times 10^{2}\\\\\rm [CO]^{2}& = 0.00326\\\\&= 0.0571\end{aligned}$

Moles of the carbon monoxide is calculated as:

$\begin{aligned}\rm n &= \rm mass \times concentration\\\\&= 3.50 \;\rm L \times 0.057 14 \;\rm mol\; L^{-1}\\\\&= 0.200\;\rm mol\end{aligned}$

Therefore, at equilibrium, the moles of the carbon monoxide is 0.200 mol.

Learn more about moles here:

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