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JEE Main levelSingle correct optionDifficulty 2 of 3ce013

For $CaCO X_{3} (s) CaO (s) + CO X_{2} (g)$ , $K_{p} = 0.80 atm$ at 1000 K. Taking $R = 0.0821 L atm mo l^{- 1} K^{- 1}$ , $K_{c}$ (in $mol L^{- 1}$ ) at this temperature is

A $6.57 \times 1 0^{1}$
B $1.19 \times 1 0^{- 4}$
C $9.74 \times 1 0^{- 3}$
D $8.00 \times 1 0^{- 1}$

Answer: C

K_{p} = K_{c} (RT)^{Δ n_{g}}

with

Δ n_{g} = 1

, so

K_{c} = K_{p} / (RT)

.

RT = 0.0821 \times 1000 = 82.1

, giving

K_{c} = 0.80/82.1 = 9.74 \times 1 0^{- 3} mol L^{- 1}

.

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