Answer:
2.27 M
Explanation:
Given that :
volume = 250.0 mL = 0.250 L
Number of moles of [tex]I_2[/tex] = 1.3 mol
Number of moles of [tex]HI[/tex] = 1.0 mole
Initial concentration of [tex]I_2[/tex] = [tex]\frac{numbers of mole}{volume}[/tex]
= [tex]\frac{1.3}{0.250}[/tex]
= 5.20 M
Initial concentration of [tex]HI[/tex] = [tex]\frac{numbers of mole}{volume}[/tex]
= [tex]\frac{1.0}{0.250}[/tex]
= 4.0 M
Equation of the reaction is represented as:
[tex]H_2}_{(g)} + I_2_{(g)}----->2HI_{(g)}[/tex]
The I.C.E Table is as follows:
[tex]H_2}_{(g)}[/tex] + [tex]I_2_{(g)}[/tex] -----> [tex]2HI_{(g)}[/tex]
Initial(M) 0.0 5.20 4.0
Change +x +x - 2x
Equilibrium(M) x 5.20+x 4 - 2x
[tex]K = \frac{[HI]^2}{[H_2][I_2]}[/tex]
[tex]K = \frac{[4-2x]^2}{[x][5.20+x]}[/tex] where K = 0.983
[tex]0.983 = \frac{(4-2x)^2}{(5.20x+x^2)}[/tex]
[tex]0.983(5.20x+x^2) = (4-2x)^2[/tex]
[tex]5.1116x + 0.983x^2=16-16x+4x^2[/tex]
[tex]5.1116x +16x + 0.983x^2 -4x^2 -16 =0[/tex]
[tex]21.1116x - 3.017x^2 -16 =0[/tex]
multiplying through by (-) and rearranging in the order of quadratic equation; we have:
[tex]3.017x^2 -21.1116x +16[/tex]
[tex]x^2 - 6.9975+5.30 =0[/tex]
using the quadratic formula:
= [tex]\frac{-b \pm\sqrt{b^2-4ac} }{2a}[/tex]
= [tex]\frac{-(-6.9975) \pm\sqrt{(-6.9975)^2-4(1)(5.3)} }{2(1)}[/tex]
= [tex]\frac{-(-6.9975) + \sqrt{(-6.9975)^2-4(1)(5.3)} }{2(1)}[/tex] OR [tex]\frac{-(-6.9975) - \sqrt{(-6.9975)^2-4(1)(5.3)} }{2(1)}[/tex]
= 6.133 OR 0.864
since 6.133 is greater than K value then it is void, so we go by the lesser value which is 0.864
so x = 0.864 M
The equilibrium molarity of HI = (4.0- 2x)
= 4.0 - 2(0.864)
= 4.0 - 1.728
= 2.272 M
= 2.27 M to two decimal places
