Asked by Tom
At 462degrees, the reaction
(1) heat + 2 NOCl(g) <=====> 2 NO(g) + Cl2(g)
has an equilibrium constant, Keq = 8.0 x 10-2.
a) What is Keq at 462degrees for the reaction
2 NO(g) + Cl2(g) <=====> 2 NOCl(g)
1/Keq
may I ask how did you get this answer?
Keq for the reverse reaction is simply the reciprocal.
ok but What if the equation was then NOCl <=> NO + 1/2Cl2 ?
Then Keq for the reaction as written is sq rt original Keq.
It works this way.
2A + B2 ==> 2C Keq = 10
A + 1/2 B2 ==> C (Keq)<sup>1/2</sup> = sq rt 10
4A + 2B2 ==> 4C (Keq)<sup>2</sup> = 100
2C ==> 2A + B2 Keq = 0.1
(1) heat + 2 NOCl(g) <=====> 2 NO(g) + Cl2(g)
has an equilibrium constant, Keq = 8.0 x 10-2.
a) What is Keq at 462degrees for the reaction
2 NO(g) + Cl2(g) <=====> 2 NOCl(g)
1/Keq
may I ask how did you get this answer?
Keq for the reverse reaction is simply the reciprocal.
ok but What if the equation was then NOCl <=> NO + 1/2Cl2 ?
Then Keq for the reaction as written is sq rt original Keq.
It works this way.
2A + B2 ==> 2C Keq = 10
A + 1/2 B2 ==> C (Keq)<sup>1/2</sup> = sq rt 10
4A + 2B2 ==> 4C (Keq)<sup>2</sup> = 100
2C ==> 2A + B2 Keq = 0.1
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