Question

An equilibrium mixture of CO, O2 and CO2 at a certain temperature contains 0.0010 M CO2 and 0.0100 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?
A) 7.1 × 10-7 M.
B) 8.4 × 10-4 M.
C) 1.4 × 10-2 M.
D) 1.2 × 10-1 M.

Answer 1

The correct answer to this question is letter "B) 8.4 × 10-4 M."

2CO+O2<->2CO2

-x       -x        +x

It started as 2CO2 so +x, then it decomposed into 2CO and O2, so-x

2CO-oxidized, O2-Oxidized, 2CO2-reduced

the easiest way is to look at the equation, so if we have some reactant and no product 

A<->2C+D

-x    +2x +x
so if we had product concentration, and no reactant concentration 3A<-> C+D +3x -x-x

Answer 2

Answer: The correct answer is Option B.

Explanation:

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{c}$

The given chemical reaction follows:

$2CO(g)+O_2(g)\rightleftharpoons 2CO_2(g)$

The expression of $K_{c$ for above reaction is:

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

We are given:

$K_c=1.4\times 10^2$

$[CO_2]=0.0010M$

$[O_2]=0.0100M$

Putting values in above equation, we get:

$1.4\times 10^2=\frac{(0.0010)^2}{[CO]^2\times (0.0100)}$

$[CO]=\sqrt{\frac{(0.0010)^2}{0.0100\times 1.4\times 10^2}}$

$[CO]=8.4\times 10^{-4}M$

Hence, the correct answer is Option B.