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Sagot :
Answer:
A(g)+2B(g)→C(g)+D(g) K
C
=10
12
At 0.5 1 0.5 0.5
t=0
At 0.5−α 1−2α 0.5+α 3.5+α
t=t
eq
Since K
C
=10
12
very large α≈0.5
Let 0.5−α be (x)
So [A]=x
[B]=x
[C]≈1
[D]≈4
K
C
=
[B]
2
[A]
[C][D]
4x
2
×x
1×4
=10
+12
x=10
−4
[B]=2x=2×10
−4
M
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