Question

An equilibrium mixture of CO, O2 and CO2 at a certain temperature contains 0.0010 M CO2 and 0.0025 M O2. At this temperature, Kc equals 1.4 × 102 for the reaction: 2 CO(g) + O2(g) ⇌ 2 CO2(g). What is the equilibrium concentration of CO?

Answer 1

Answer:

$5.35 *10^{-4}$M

Explanation:

Equation for the reaction is as follows:

$2CO_{(g)}$      +      $O_{2(g)}$        ⇄       $2CO_{2(g)}$

By Applying the ICE Table; we have

                             $2CO_{(g)}$      +      $O_{2(g)}$        ⇄       $2CO_{2(g)}$

Initial                      x                  0.0025 M              0.0010 M

Change                  0                       0                            0

Equilibrium             x                  0.0025 M              0.0010 M

$K_c =\frac{[CO_2]^2}{[CO]^2[O_2]}$

Given that $K_c = 1.4*10^2$ ; Then:

$1.4 *10^2 = \frac{(0.001)^2}{(x)^2(0.025)}$

$1.4 *10^2*0.025 = \frac{(0.001)^2}{(x)^2}$

$3.5 =( \frac{(0.001)}{(x)})^2$

$\sqrt {3.5} = \sqrt {( \frac{(0.001)}{(x)} )^2}$

$1.87=\frac{(0.001)}{(x)}$

$(x)= \frac{(0.001)}{1.87 }$

$x = 5.35 *10^{-4}$M

∴ The equilibrium concentration of CO = $x = 5.35 *10^{-4}$M