17. At 700 K, the equilibrium constant, K , for the reaction
2SO3(9) %2502(g) + O2 (g) is 1.8 * 10-3 atm. The
value of K. for the above reaction at the same
temperature in moles per litre would be
(1) 1.1 x 10-7
(2) 3.1 x 10-5
(3) 6.2 x 10-7
(4) 9.3 x 10-7
Answers
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Answer:
Given: 2SO
3
⇌2SO
2
+O
2
1 atm=10
5
pa
Δn=n
products
−n
reactants
K
p
=1.8×10
−3
KPa=1.8Pa=1.8×10
−5
atm
T=700K
R=0.0821lit−atm
K
p
=K
c
(RT)
Δn
g
K
c
=
RT
K
p
K
c
=
700×0.0821
1.8×10
−5
=0.031×10
−5
=3.1×10
−7
molL
−1
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