Question #fdae0
1 Answer
Explanation:
For starters, notice that
Now, since you know how to set up the ICE table, I'll just walk you through what's happening without making an ICE table of my own.
The equilibrium reaction given to you is
#"A " + " B " rightleftharpoons " C " + " D"#
Notice that all the chemical species are present in
#["A"] = "2.00 M" - x -># the concentration of#"A"# decreased by#x#
#["B"] = "2.00 M" - x -># the concentration of#"B"# decreased by#x#
#["C"] = 0 + x -># the concentration of#"C"# increased by#x#
#["D"] = 0 + x -># the concentration of#"D"# increased by#x#
Now all you have to do is plug these values into the expression of the equilibrium constant,
#K_c = (["C"] * ["D"])/(["A"] * ["B"])#
You will end up with
#K_c = (x * x)/((2.00 - x)(2.00 - x))#
#3.5 = x^2/(2.00 - x)^2#
Rearrange this to quadratic equation form
#3.5 * (2.00 - x)^2 = x^2#
#2.5x^2 - 14x + 14 = 0#
This quadratic equation will produce two positive values
#{(x_1 = 4.30), (x_2 = 1.30) :}#
Since the equilibrium concentrations of all species involved in the reaction must be
#x = 1.30#
This means that the equilibrium concentration of
#color(green)(bar(ul(|color(white)(a/a)color(black)(["A"] = "2.00 M" - "1.3 M" = "0.70 M")color(white)(a/a)|)))#