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JEE Advanced levelSingle correct optionDifficulty 3 of 3ce091

For $A (g) 2 B (g)$ , $K_{c} = 0.0250 mol d m^{- 3}$ at 600 K. Using $R = 8.314 J mo l^{- 1} K^{- 1}$ , $1 d m^{3} = 1 0^{- 3} m^{3}$ and $1 bar = 1 0^{5} Pa$ , $K_{p}$ in bar is

A $1.25 \times 1 0^{- 3}$
B $1.25$
C $124.7$
D $1.25 \times 1 0^{5}$

Answer: B

Convert

K_{c}

to SI:

0.0250 mol d m^{- 3} = 25.0 mol m^{- 3}

. With

Δ n = + 1

,

K_{p} = K_{c} RT = 25.0 \times 8.314 \times 600 = 124710 Pa

(since

J m^{- 3} = Pa

). Dividing by

1 0^{5}

gives 1.25 bar.

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