0.010 mol of CaCO3(s) is placed in an evacuated 5.0 L rigid vessel and heated to 1000 K, where K_p = 0.50 atm for CaCO3(s) CaO(s) + CO2(g). Take R = 0.0821 L atm mol^-1 K^-1. The final pressure in the vessel is

JEE Main levelSingle correct optionDifficulty 2 of 3ce078

0.010 mol of $CaCO X_{3} (s)$ is placed in an evacuated 5.0 L rigid vessel and heated to 1000 K, where $K_{p} = 0.50$ atm for $CaCO X_{3} (s) CaO (s) + CO X_{2} (g)$ . Take $R = 0.0821 L atm mo l^{- 1} K^{- 1}$ . The final pressure in the vessel is

To reach

p_{CO X_{2}} = K_{p} = 0.50

atm the vessel needs

n = p V / RT = 0.50 \times 5.0/ (0.0821 \times 1000) = 0.0305

mol of

CO X_{2}

. Only 0.010 mol is available, so all the

CaCO X_{3}

decomposes and equilibrium is never reached. Then

p = n RT / V = 0.010 \times 82.1/5.0 = 0.164

atm.