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Gaseous hydrogen and gaseous iodine react together to form hydrogen iodide. H2 + I2 ⇌ 2HI The graph shows how the amount of hydrogen iodide varies with time in a 1.00 dm3 container. The initial amounts of hydrogen and iodine were 1.00 mol H2 and 1.00 mol I2.

enter image description here

Calculate the value of Kc

My attempt:

As the equilibrium amount of HI is 1.5 mol according to the graph, through stoichiometry the equilibrium amounts of H2 and I2 are 0.25 mol each. Since the volume is given as 1 dm3 the equilibrium concentrations would be [HI] = 1.5, [H2]=[I2]=0.25

Therefore I calculate Kc = 1.5/(0.25 x 0.25) = 24

The answer for Kc is given as 9 and the equilibrium amounts for H2 and I2 are shown to be correct however the concentration for HI is given as 0.75. Kc = (0.75)^2/(0.25 x 0.25) = 9

I don't see how this answer comes about.

Since the total amount of HI at equilibrium is given as 1.5, if I divide 1.5 by 2 as there are 2 moles of HI produced I would get the mentioned 0.75 as the [HI] at equilibrium, but that doesn't make sense since that would mean there's 0.75^2 = 0.5625 mol at equilibrium, which is not shown in the graph.

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  • $\begingroup$ You used the formula for Kc wrong, its Kc = [HI]^2/[I2][H2], however that still doesn't give the required answer $\endgroup$ Apr 2 at 13:21
  • $\begingroup$ Yes, I used that formula, I took the total HI concentration as 1.5 because it is the amount we get directly from the graph and didn't square it. $\endgroup$
    – Jane902
    Apr 2 at 13:35
  • $\begingroup$ No, when you calculated 24, you didn't square [HI] $\endgroup$ Apr 3 at 7:02
  • $\begingroup$ In cases the expected answer is an incorrect answer, there are no good ways how to reach that answer, but trying to simulate the incorrect thinking. // They must have forgotten to involve the factor 2, forming 2 HI from 1 I2 + 1 H2. $\endgroup$
    – Poutnik
    Apr 3 at 7:15

1 Answer 1

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Let:

A = $\ce{H2}$

B = $\ce{I2}$

C = $\ce{HI}$

At equilibrium:

$$C_A=1-x$$ $$C_B=1-x$$ $$C_C=2x=1.5\implies x=0.75$$

Calculating $K_c$:

$$K_c=\frac{C_C^2}{C_AC_B}=\frac{\left(2x\right)^2}{\left(1-x\right)^2}=\left(\frac{2x}{1-x}\right)^2=\left(\frac{1.5}{0.25}\right)^2=36$$

I don't believe the answer given in the problem statement is correct, since this reaction exhibits no change in moles between products and reactants:

$\Delta n=2-2=0$

This means that the total equilibrium concentration $C$ must be equal to the total initial concentration $C_o$:

$$C=C_o+\require{cancel}\cancel{\Delta nx}\implies C=C_o$$

If we use the values I provided:

$$C=C_A+C_B+C_C=\left(0.25+0.25+1.5\right)\pu{mol/L}=\pu{2mol/L}$$

$$C_o=\left(1+1\right)\pu{mol/L}=\pu{2mol/L}$$

So the condition: $C=Co$ is satisfied.

If however, we use the values assumed by the answer given:

$$C=C_A+C_B+C_C=\left(0.25+0.25+0.75\right)\pu{mol/L}=\pu{1.25mol/L}$$

$$C_o=\left(1+1\right)\pu{mol/L}=\pu{2mol/L}$$

We get: $C\neq C_o$, so that can't be correct.

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