Thermal decomposition of gaseous $X_2$ to gaseous $X$ at $298 \ K$ takes place according to the following equation :
$X_{2(g)} \rightleftharpoons 2 X_{(g)}$
The standard reaction Gibbs energy,$\Delta_r G^{\circ}$,of this reaction is positive. At the start of the reaction,there is one mole of $X_2$ and no $X$. As the reaction proceeds,the number of moles of $X$ formed is given by $\beta$. Thus,$\beta_{\text{equilibrium}}$ 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 : $R=0.083 \ L \ bar \ K^{-1} \ mol^{-1}$)
$(1)$ The equilibrium constant $K_P$ for this reaction at $298 \ K$,in terms of $\beta_{\text{equilibrium}}$,is
$(A)$ $\frac{8 \beta_{\text{equilibrium}}^2}{2-\beta_{\text{equilibrium}}}$ $(B)$ $\frac{8 \beta_{\text{equilibrium}}^2}{4-\beta_{\text{equilibrium}}^2}$ $(C)$ $\frac{4 \beta_{\text{equilibrium}}^2}{2-\beta_{\text{equilibrium}}}$ $(D)$ $\frac{4 \beta_{\text{equilibrium}}^2}{4-\beta_{\text{equilibrium}}^2}$
$(2)$ 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 $X_2$ takes place spontaneously
$(C)$ $\beta_{\text{equilibrium}}=0.7$
$(D)$ $K_c < 1$
$X_{2(g)} \rightleftharpoons 2 X_{(g)}$
The standard reaction Gibbs energy,$\Delta_r G^{\circ}$,of this reaction is positive. At the start of the reaction,there is one mole of $X_2$ and no $X$. As the reaction proceeds,the number of moles of $X$ formed is given by $\beta$. Thus,$\beta_{\text{equilibrium}}$ 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 : $R=0.083 \ L \ bar \ K^{-1} \ mol^{-1}$)
$(1)$ The equilibrium constant $K_P$ for this reaction at $298 \ K$,in terms of $\beta_{\text{equilibrium}}$,is
$(A)$ $\frac{8 \beta_{\text{equilibrium}}^2}{2-\beta_{\text{equilibrium}}}$ $(B)$ $\frac{8 \beta_{\text{equilibrium}}^2}{4-\beta_{\text{equilibrium}}^2}$ $(C)$ $\frac{4 \beta_{\text{equilibrium}}^2}{2-\beta_{\text{equilibrium}}}$ $(D)$ $\frac{4 \beta_{\text{equilibrium}}^2}{4-\beta_{\text{equilibrium}}^2}$
$(2)$ 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 $X_2$ takes place spontaneously
$(C)$ $\beta_{\text{equilibrium}}=0.7$
$(D)$ $K_c < 1$
- A
- B
- C
- D
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Similar Questions
For the reaction,$NO_2 + CO \rightleftharpoons NO + CO_2$,one mole of $NO_2$ and $2$ moles of $CO$ were kept in a vessel. Calculate the equilibrium constant $K_p$,if at equilibrium $25 \%$ of the initial amount of $CO$ is consumed.
In a closed vessel,the decomposition of solid $Mg(HCO_3)_2$ is given as follows:
$Mg(HCO_3)_{2(s)} \rightleftharpoons MgCO_{3(s)} + CO_{2(g)} + H_2O_{(g)}$
Given $K_p = 64 \ bar^2$ and the mole fraction of $CO_2$ is $0.8$,calculate the total pressure at equilibrium. (in $bar$)
$Mg(HCO_3)_{2(s)} \rightleftharpoons MgCO_{3(s)} + CO_{2(g)} + H_2O_{(g)}$
Given $K_p = 64 \ bar^2$ and the mole fraction of $CO_2$ is $0.8$,calculate the total pressure at equilibrium. (in $bar$)
Equilibrium constants for the following reactions are given (pressure in $atm$) for the reactions at $0\,^{\circ}C$. Select the option mentioning the correct order of True $(T)$ or False $(F)$ statements:
$(A) \ A \cdot 6H_2O_{(s)} \rightleftharpoons A \cdot 2H_2O_{(s)} + 4H_2O_{(g)}; \ K_P = 1.6 \times 10^{-11}$
$(B) \ B \cdot 12H_2O_{(s)} \rightleftharpoons B \cdot 7H_2O_{(s)} + 5H_2O_{(g)}; \ K_P = 2.43 \times 10^{-13}$
$(C) \ C \cdot 10H_2O_{(s)} \rightleftharpoons C_{(s)} + 10H_2O_{(g)}; \ K_P = 10^{-30}$
Aqueous tension of $H_2O$ at $0\,^{\circ}C$ is given as $0.76 \ torr$.
$(I)$ The most effective drying agent will be $C_{(s)}$ out of $C_{(s)}$,$B \cdot 7H_2O_{(s)}$,and $A \cdot 2H_2O_{(s)}$.
$(II)$ At $0\,^{\circ}C$,$A \cdot 6H_2O_{(s)}$ and $B \cdot 12H_2O_{(s)}$ will be efflorescent.
$(III)$ If $R.H.$ is less than $100\%$ in a chamber at $0\,^{\circ}C$,then none of the substances can act as deliquescent.
$(A) \ A \cdot 6H_2O_{(s)} \rightleftharpoons A \cdot 2H_2O_{(s)} + 4H_2O_{(g)}; \ K_P = 1.6 \times 10^{-11}$
$(B) \ B \cdot 12H_2O_{(s)} \rightleftharpoons B \cdot 7H_2O_{(s)} + 5H_2O_{(g)}; \ K_P = 2.43 \times 10^{-13}$
$(C) \ C \cdot 10H_2O_{(s)} \rightleftharpoons C_{(s)} + 10H_2O_{(g)}; \ K_P = 10^{-30}$
Aqueous tension of $H_2O$ at $0\,^{\circ}C$ is given as $0.76 \ torr$.
$(I)$ The most effective drying agent will be $C_{(s)}$ out of $C_{(s)}$,$B \cdot 7H_2O_{(s)}$,and $A \cdot 2H_2O_{(s)}$.
$(II)$ At $0\,^{\circ}C$,$A \cdot 6H_2O_{(s)}$ and $B \cdot 12H_2O_{(s)}$ will be efflorescent.
$(III)$ If $R.H.$ is less than $100\%$ in a chamber at $0\,^{\circ}C$,then none of the substances can act as deliquescent.
For the equilibrium $ \text{Liquid} \rightleftharpoons \text{Vapour} $,which of the following expressions is correct?
Easy
View SolutionFor the reaction $CH_4(g) + 2O_2(g) \rightleftharpoons CO_2(g) + 2H_2O(l)$,the enthalpy change is $\Delta H_r = -170.8 \, kJ \, mol^{-1}$. Which of the following statements is $NOT$ true?
Medium
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