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Ideal Gas Equation

The process ...

Question

The process $C a {C O}_{3} ⇌ C a O + {C O}_{2}$ attains equilirbium in $11.2$ lit vessel at $1092 K$ and $2$ atm. Now, initial number of moles of $C A {C O}_{3}$ cannot be?

1/2

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1/3

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1/4

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2/3

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Solution

The correct option is C 1/4
The equation at its equilibrium,
${C a C O}_{3} ⇋ C a O + {C O}_{2}$

$K_{P} = P_{{C O}_{2}}$

$P_{{C O}_{2}} = 5 a t m$
Equilibrium constant is same.

The moles of ${C O}_{2},$ $n_{{C O}_{2}}$

$= \frac{P V}{R T} = \frac{2 \times 11.2}{0.082 \times 1092}$

$= 0.25 = \frac{1}{4}$

At equilibrium moles of ${C a C O}_{2}$ $=$ Moles of ${C O}_{2}$

The initial moles of ${C a C O}_{3}$ $= \frac{1}{4}$

Hence the moles of ${C a C O}_{3}$ can not be $\frac{1}{2}, \frac{1}{3}, \frac{2}{3}$

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The process CaCO3⇌CaO+CO2 attains equilirbium in 11.2 lit vessel at 1092K and 2 atm. Now, initial number of moles of CACO3 cannot be?

The correct option is C 1/4The equation at its equilibrium,CaCO3⇋CaO+CO2KP=PCO2PCO2=5 atmEquilibrium constant is same.The moles of CO2, nCO2=PVRT=2×11.20.082×1092=0.25=14At equilibrium moles of CaCO2 = Moles ofCO2 The initial moles of CaCO3 =14Hence the moles ofCaCO3 can not be 12,13,23

The process $C a {C O}_{3} ⇌ C a O + {C O}_{2}$ attains equilirbium in $11.2$ lit vessel at $1092 K$ and $2$ atm. Now, initial number of moles of $C A {C O}_{3}$ cannot be?