AL questions

2009-09-30 6:26 am

1. I have tried hard to solve this problem but to no avail.

Equal amounts of hydrogen and iodine are allowed to reach an
equilirium at 298K:

H2(g)+ I2(g)≒2HI(g)

where ≒ means backward and forward reactions.

If 80% of hydrogen is converted to hydrogen iodide at the
equilibrium, what is the value of Kp at this temperature?

2. What is the difference between mesomeric effect and resonance
effect?

回答 (1)

六呎將軍

2009-09-30 6:48 am

✔ 最佳答案

(1) Suppose that the initial partial pressures of H2 and I2 are P, then from the equation, equimolar of H2 and I2 react to give a double mole no. of HI, so at equilibrium:
Partial pressure of H2 = 0.2P
Partial pressure of I2 = 0.2P
Partial pressure of HI = 1.6P
So Kp = (1.6P)2/(0.2P x 0.2P)
= 64 (No unit)
(2) There's no difference between the 2 effects. We can state that mesomeric effect is a form of resonance, giving different structures for stabilizing the molecule.

2009-10-01 15:10:18 補充：
Since 80% of hydrogen is converted, 20% is remained. Thus 0.2 arises.

2009-10-02 17:31:02 補充：
1. You are welcome.
2. Yes, by Dalton's law, mole ratio is equivalent to partial pressure ratio
3. In fact NOT all iodine has been converted. Like hydrogen, only 80% of it has been converted.
4. 298K is just to indicate that it is a standard condition.

參考: Myself

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