Q. The thermal dissociation equilibrium of $\mathrm{CaCO}_{3}(\mathrm{s})$ is studied under different conditions.
$\mathrm{CaCO}_{3}(\mathrm{s}) \square \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})$
For this equilibrium, the correct statement(s) is(are)
(A) $\Delta \mathrm{H}$ is dependent on $\mathrm{T}$
(B) $\mathrm{K}$ is independent of the initial amount of $\mathrm{CaCO}_{3}$
(C) $\mathrm{K}$ is dependent on the pressure of $\mathrm{CO}_{2}$ at a given $\mathrm{T}$
(D) $\Delta \mathrm{H}$ is independent of the catalyst, if any
[JEE 2013]
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Sol. (A,B,D)
s independent of catalyst but dependent on temperature
Paragraph 1 (2 TO 3) Thermal decomposition of gaseous $\mathrm{X}_{2}$ to gaseous X at 298 K takes place according to the following equation :
The standard reaction Gibbs energy, 
, of this reaction is positive. At the start of the
reaction, there is one mole of X2 and no X. As the reaction proceeds, the number of moles of
X formed is given 
is the number of moles of X formed at equilibrium. The
reaction is carried out at a constant total pressure of 2
bar. Consider the gases to behave ideally.
(Given $\left.: \mathrm{R}=0.083 \mathrm{L} \text { bar } \mathrm{K}^{-1} \mathrm{mol}^{-1}\right)$
Q. The equilibrium constant KP for this reaction at 298 K, in terms of 
, is
[JEE – Adv. 2016]
, is
[JEE – Adv. 2016]
Q. The INCORRECT statement among the following , for this reaction, is
(A) Decrease in the total pressure will result in formation of more moles of gaseous X
(B) At the start of the reaction, dissociation of gaseous $\mathrm{X}_{2}$ takes place spontaneously
(C) $\beta_{\text {equilibrium }}=0.7$
(D) $\mathrm{K}_{\mathrm{C}}<1$
[JEE – Adv. 2016]
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Sol. (C) (A) On decreasing $\mathrm{P}_{\mathrm{T}}\left[\mathrm{Q}=\frac{\mathrm{n}_{x} \mathrm{P}_{\mathrm{T}}}{\mathrm{n}_{\mathrm{x}_{2}} \mathrm{n}_{\mathrm{T}}}\right]$ Q will be less than Kp reaction will move in forward direction (B) At the start of the reaction $\Delta \mathrm{G}=\Delta \mathrm{G}^{0}+\mathrm{RT} \ln \mathrm{Q}$ $\mathrm{t}=0, \mathrm{Q}=0 \Rightarrow \Delta_{\mathrm{rxn}} \mathrm{G}=-\mathrm{ve} \quad$ (spontaneous)
Q. The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K are
$\Delta_{f} \mathrm{G}^{\circ}[\mathrm{C}(\text { graphite })]=0 \mathrm{kJ} \operatorname{mol}^{-1} \Delta_{f} \mathrm{G}^{\circ}[\mathrm{C}(\text { diamond })]=2.9 \mathrm{kJ} \mathrm{mol}^{-1}$
The standard state means that the pressure should be 1 bar , and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by $2 \times 10^{-6} \mathrm{m}^{3} \mathrm{mol}^{-1}$. If C(graphite) is converted to C(diamond) isothermally at T = 298 K, the pressure at which C(graphite) is in equilibrium with C(diamond), is
[Useful information : $\left.1 \mathrm{J}=1 \mathrm{kg} \mathrm{m}^{2} \mathrm{s}^{-2} ; 1 \mathrm{Pa}=1 \mathrm{kg} \mathrm{m}^{-1} \mathrm{s}^{-2} ; 1 \text { bar }=10^{5} \mathrm{Pa}\right]$
(A) 14501 bar (B) 29001 bar (C) 58001 bar (D) 1405 bar
[JEE – Adv. 2017]
Nice questions
Thanks esaral
some nice stuff!!!