Question

How can I calculate `K_c` of this reaction?

Answer 1

`"1.33 L/mmol"`

Explanation:

First, find out reactants and products from given graph.

At `"t = 0"` concentration of `"C"` is zero. So, it should be product, `"A"` and `"B"` should be the reactants.

`"A + B " rightleftharpoons " C"`

`"K"_"c" = (["C"])/(["A"]["B"])`

Where, `["A"], ["B"]` and `["C"]` are concentration of `"A", "B"` and `"C"` at equilibrium respectively.

At equilibrium,

`["A"] = "0.5 mmol/L"`

`["B"] = "1.5 mmol/L"`

`["C"] = "1 mmol/L"`

`"K"_"c" = "1 mmol/L"/("0.5 mmol/L × 1.5 mmol/L") = 4/3"L/mmol" = "1.33 L/mmol"`

Answer 2

The value of `K_text(c)` depends on the equilibrium reaction.

From the graph, the equilibrium concentrations are

`["A"] = "0.5 mmol/L"; ["B"] = "1.5 mmol/L"; ["C"] = "1.0 mmol/L"`

The general form of an equilibrium constant expression is

`K_text(c) = (["Products"])/(["Reactants"])`

The concentrations should be in moles per litre, but your instructor appears to be using millimoles per litre.

If the reaction is `"A + B ⇌ C"`

`K_text(c) = (["C"])/(["A"]["B"]) = 1/(0.5 × 1.5) = 1.3`

If the reaction is `"2A + B ⇌ C"`

`K_text(c) = (["C"])/(["A"]^2["B"]) = 1/(0.5^2 × 1.5) = 2.7`

If the reaction is `"A + 2B ⇌ C"`

`K_text(c) = (["C"])/(["A"]["B"]^2) = 1/(0.5 × 1.5^2) = 0.89`

Note 1: The values of equilibrium constants are dimensionless.

Note2: The answer can have only two significant figures because that is all we can read from the graphs.