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: B

Answer 1

with this one i know how to use the ICE rule, but what i dont understand is for example, when is the reactants -x and the products +x

Answer 2

kk 1 sec

Answer 3

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

Answer 4

okay i understand that
but what would make it like this

+x +x -x

Answer 5

1sec

Answer 6

kk you would need to see which is oxidized and which was reduced

Answer 7

example CO would be -, while CO2 is +

Answer 8

how do you tell whic is oxidized or kan you use the q>k q<k and if i could use those how do i use it

Answer 9

CO+H20<->CO2+H2
-x -x +x +x
oxidiation is loss -, reduction is gain+
more - more oxidation, more H or + more reduction, oilrig mnemonic

Answer 10

1 sec let me refresh on q>k rules

Answer 11

ok

Answer 12

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

so greater concentration will have +, less will be -

Answer 13

hope it makes sense

Answer 14

yes it does i just had to look at it. okay i have another question.

Answer 15

kk awseome!

Answer 16

kk