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JEE Advanced levelSingle correct optionDifficulty 3 of 3sc035

Gas A alone on a surface at its partial pressure gives $θ_{A} = 0.50$ (Langmuir). Gas B is now added, adsorbing on the same sites with $K_{B} p_{B} = 3.0$ while $p_{A}$ is unchanged. Using competitive Langmuir adsorption, the new $θ_{A}$ is:

A $0.20$
B $0.25$
C $0.50$
D $0.33$

Answer: A

Alone,

θ_{A} = K_{A} p_{A} / (1 + K_{A} p_{A}) = 0.50

gives

K_{A} p_{A} = 1.0

. With B:

θ_{A} = K_{A} p_{A} / (1 + K_{A} p_{A} + K_{B} p_{B}) = 1.0/ (1 + 1.0 + 3.0) = 0.20

. Ignoring A in the denominator gives

0.25

.

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