[Big-Daddy]

This website keeps giving me answers I can't agree with. http://www.askiitians.com/iit-jee-chemical-equilibrium/solved-problems-of-chemical-equilibrium/

Question 3:
When 3.06g of solid NH4HS is introduced into a 2 litre evacuated flask at 27°C, 30% of the solid decomposed into gaseous ammonia and hydrogen sulphide.

(i) Calculate Kc for reaction at 27°C.
(ii) What would happen to the equilibrium when more solid NH4HS is introduced into the flask?

My Solution:

(i)

Kc=[NH3]*[H2S]
30% fraction of NH4HS is dissociated thus [NH3] and [H2S] both = 0.3*[NH4HS]_t=0.

Kc=(0.3*[NH4HS]_t=0)*(0.3*[NH4HS]_t=0)

[NH4HS]_t=0=(3.06/(14+1.008*5+32.066))/2 = 0.0299377764 moldm-3

Kc=(0.3*0.0299377764)²=8.0664*10^-5 mol²dm^-6.

The website's solution:

                  NH4HS(s)  NH3(g) + H2S (g)
 
 
 
                  0.06 (1 – x)        0.06x           0.06x
 
 
 
                  as x = 30%, so, x = 0.3
 
 
 
                  So, 0.06 × 0.7    0.06´0.3 0    0.6 × 0.3
 
 
 
                  So, Kc = [NH3][H2S]
 
 
 
                  0.018 × 0.018 = 3.24 × 10–2


To me it seems the website has neglected the need to divide by 2 to take the number of moles to the concentration (given that Concentration = Moles / Volume) as the volume is not 1 dm3 but 2 dm3. Am I correct here?

And problem 1 on that website ... I cannot see how it is possible for equilibrium to produce 10/(126.9+1.008)=0.078181 moles of HI when the original moles of I2 were only 5.2/(126.9*2)=0.0204886 moles (although HI is produced twice as fast as I2 is lost, that should be enough for a maximum of 0.0204886*2 moles of HI and there are apparently more than that).

[Borek]

You are right about the being wrong, but your approach - with concentration of NH₄S(s) - doesn't make sense. Following this line of thinking if you change initial amount of the solid you change its initial concentration, so you shift the final equilibrium position.

[Big-Daddy]

You are right about the being wrong, but your approach - with concentration of NH₄S(s) - doesn't make sense. Following this line of thinking if you change initial amount of the solid you change its initial concentration, so you shift the final equilibrium position.

Hmm you are right. I had forgotten that solids are neither part of Kc expressions nor part of Q (position of equilibrium) expressions. So how would I solve that problem properly then?

[Borek]

Do you really need [NH₄S]_t=0 for anything?

[Big-Daddy]

Do you really need [NH₄S]_t=0 for anything?

I suppose that if you have Kc=(0.3*[NH4HS]t=0)*(0.3*[NH4HS]t=0) and because NH4HS is solid you can treat it as 1, Kc=0.3*0.3=0.09 mol²dm_-6 in this case?

[Borek]

How many moles of H₂S (or NH₃) are present in the gas phase?

[Big-Daddy]

How many moles of H₂S (or NH₃) are present in the gas phase?

0.3*[NH4HS]_t=0 (in this particular case) = 0.3*0.0299377764 = 8.981*10^-3 moldm^-3

No? But then the Kc I originally got does provide the right answer ... and I can't see how that can be true (because if more of the NH4HS is added, then either the equilibrium constant will change, which concentration changes cannot make it do anyway, or the fraction 0.3 will change to maintain the equilibrium constant at its value, but if it did that would change the position of equilibrium).