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Spud93
Feb 6, 2011, 06:51 AM
When hydrogen gas and iodine gas are mixed in a vessel at constant
temperature, a reaction occurs which, after a certain amount of time, reaches
equilibrium giving a mixture of the reactants and a product (hydrogen iodide).
The balanced chemical equation for this reaction is:
H2(g)+I2(g)= 2HI(g) (2)

(a)
(I) Write down the equilibrium expression for Reaction 2.
(ii) For the equilibrium reaction between hydrogen and iodine, apply
Le Chatelier's principle to explain what effect increasing and
decreasing the pressure will have on the equilibrium yield of hydrogen
iodide in Reaction 2. (We expect that you will be able to answer part
(a) (ii) in about 100 words.)
(b) Table 2 gives the results of several experiments for Reaction 2 where three
different mixtures of H2
and I2
gases were allowed to react and reach
equilibrium at a temperature of 427 °C. From the data in Table 2, and
using your equilibrium expression from part (a) (I), calculate the average
value of the equilibrium constant, K, for Reaction 2 at 427 °C. Give all
your values for K, including the final average value of K, to one decimal
place.
Table 2 Concentrations of H2, I2 and HI gases at equilibrium for three
different mixtures at 427 °C
Experiment::::[H2]/mol dm−3::::[I2]/mol dm−3::::[HI]/mol dm−3
1 ::::4.80 × 10−4 ::::4.80 × 10−4 ::::3.53 × 10−3
2 ::::1.14 × 10−3 ::::1.14 × 10−3 ::::8.41 × 10−3
3 ::::4.56 × 10−3 ::::7.40 × 10−4 ::::1.35 × 10−2
(c) Table 3 gives values for the equilibrium constant at several temperatures.
The missing value of K at 427 °C is one you have determined in part (b).
(I) From the data in Table 3, how does the value of the equilibrium
constant, K, change with respect to temperature? Does the value you
determined in part (b) fit with this trend? (A couple of sentences) You
do not need to copy out Table 3 in your answer.
(ii) Would you expect this reaction to be exothermic or endothermic? Give
brief reasoning of how you have arrived at your answer in light of the
data given in Table 3. (Three or four sentences)
Table 3 The equilibrium constant, K, for Reaction 2 as a function of
temperature

T/^oC K
25 794.4
227 128.3
327 81.1
427
527 43.6
627 35.7

Unknown008
Feb 6, 2011, 10:26 AM
It would be good if you post your attempt at the question. Part (a) can normally be found in notes or in your textbook, part (b) is just a matter of plugging in values and getting an average. Part (c) requires you to analyse the data you got and make deductions.

Spud93
Feb 7, 2011, 05:05 AM
4a.I) K = (HI)^2 / (H2) (I2)
ii) Increasing the pressure will favor the side with fewer mols and the and the opposite for decreasing pressure. However, since we have two mols of reactants and two mols of product changing the pressure will have no effect on the equilibrium position (this is a shortened version of my answer) as for b. an c. I'm stuck :(

Unknown008
Feb 7, 2011, 06:14 AM
a) (i) Actually, what I would have put is:

K_c = \frac{[HI]^2}{[H_2][I_2]}

since we are dealing with gases and when we do so, the equation is in terms of partial pressure.

(ii) is okay, though I'd have liked to see your full answer.

b)

Use the first row and calculate K_c

Here's the first one:

K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{[3.53\times10^{-3}]^2}{[4.80\times10^{-4}][4.80\times10^{-4}]} = 54.1

Can you do the rest now?