1hey ne1 is not able 2 solve
Kp is 0.0529 at 26°C for the react.
NH4HS===NH3 +H2S
0.092 mole of solid NH4HS is introduced into 2.46L of evacuated flask at 26°C
A) cal. % of solid NH4HS decomposed
B) find the no. of moles of NH3 to be added to reduce the decomposition of 1% solid at the same temp.
Ans.
A)25.08%
B)0.5779
hint your antilog & log table with calulator too
give solution..
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4 Answers
1357assume that α moles got decomposed
P=αRT/V for both NH3 and H2S
hence... P2=.0529
P=√.0529=0.23
you have not given the unit of Kp... I am assuming it based on atm pressure.
α=.23x2.46/(.82x299) = .023 moles
% decom = .0023/.092*100 = 25%
1357second part you have to take
Kp=α(α+n) we know the new alpha 1% of the initial no of moles and n is someting we have to find...
1thanx bhiaya i was taking wrong in part-2 taking 1% of original alpha thanx once again but plz reply 2 quest. of day