If two gases AB2 and B2C are mixed the following equilibria are readily established AB2(g)+B2C(g)→AB3(g)+BC(g) BC(g)+B2C(g)→B3C2(g) If the reaction is started only with AB2 with B2C, then which of the following is necessarily true at equilibrium:
[AB3]eq=[BC]eq
[AB3]eq>[BC]eq
[AB3]eq>[B3C2]eq
[AB2]eq=[B2C]eq
A
[AB3]eq=[BC]eq
B
[AB2]eq=[B2C]eq
C
[AB3]eq>[B3C2]eq
D
[AB3]eq>[BC]eq
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Solution
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Let reactions is started with a mole of AB2 and b mole of B2C ⇒AB2(g)+B2C(g)→AB3(g)+BC(g) a b 0 0 a - x b - x - y x x - y BC(g)+B2C(g)→B2C2(g) x - y b - x - y y As x > y Clearly [AB3]eq>[B3C2]eq and [AB3]eq>[BC]eq
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