For the reaction $X(s) \rightleftharpoons Y(s) + Z(g)$,the plot of $\ln \frac{p_z}{p^\ominus}$ versus $\frac{10^4}{T}$ is given below,where $p_z$ is the pressure (in bar) of the gas $Z$ at temperature $T$ and $p^\ominus = 1 \ bar$.
(Given,$\frac{d(\ln K)}{d(\frac{1}{T})} = -\frac{\Delta H^\ominus}{R}$,where the equilibrium constant,$K = \frac{p_z}{p^\ominus}$ and the gas constant,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$)
$(1)$ The value of standard enthalpy,$\Delta H^\ominus$ (in $kJ \ mol^{-1}$) for the reaction is. . . . . . .
$(2)$ The value of $\Delta S^\ominus$ (in $J \ K^{-1} \ mol^{-1}$) for the given reaction,at $1000 \ K$ is. . . . . .
Give the answer for $(1)$ and $(2)$
(Given,$\frac{d(\ln K)}{d(\frac{1}{T})} = -\frac{\Delta H^\ominus}{R}$,where the equilibrium constant,$K = \frac{p_z}{p^\ominus}$ and the gas constant,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$)
$(1)$ The value of standard enthalpy,$\Delta H^\ominus$ (in $kJ \ mol^{-1}$) for the reaction is. . . . . . .
$(2)$ The value of $\Delta S^\ominus$ (in $J \ K^{-1} \ mol^{-1}$) for the given reaction,at $1000 \ K$ is. . . . . .
Give the answer for $(1)$ and $(2)$
- A$164.28, 141.32$
- B$166.28, 141.33$
- C$160.28, 141.35$
- D$166.28, 141.34$
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Similar Questions
Which of the following statements is/are true about equilibrium?
$(a)$ Equilibrium is possible only in a closed system at a given temperature.
$(b)$ All the measurable properties of the system remain constant at equilibrium.
$(c)$ Equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the reaction in the forward direction.
$(a)$ Equilibrium is possible only in a closed system at a given temperature.
$(b)$ All the measurable properties of the system remain constant at equilibrium.
$(c)$ Equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the reaction in the forward direction.
MediumKCET 2025
View SolutionExplain the relationship between the time taken to reach equilibrium and thermodynamics.
Medium
View SolutionFor a reaction,$A \rightleftharpoons P$,the plots of $[A]$ and $[P]$ with time at temperatures $T_1$ and $T_2$ are given below. If $T_2 > T_1$,the correct statement$(s)$ is (are) (Assume $\Delta H^{\ominus}$ and $\Delta S^{\ominus}$ are independent of temperature and ratio of $\ln K$ at $T_1$ to $\ln K$ at $T_2$ is greater than $T_2 / T_1$. Here $H, S, G$ and $K$ are enthalpy,entropy,Gibbs energy and equilibrium constant,respectively.)
$(A)$ $\Delta H^{\ominus} < 0, \Delta S^{\ominus} < 0$
$(B)$ $\Delta G^{\ominus} < 0, \Delta H^{\ominus} > 0$
$(C)$ $\Delta G^{\ominus} < 0, \Delta S^{\ominus} < 0$
$(D)$ $\Delta G^{\ominus} < 0, \Delta S^{\ominus} > 0$
$(A)$ $\Delta H^{\ominus} < 0, \Delta S^{\ominus} < 0$
$(B)$ $\Delta G^{\ominus} < 0, \Delta H^{\ominus} > 0$
$(C)$ $\Delta G^{\ominus} < 0, \Delta S^{\ominus} < 0$
$(D)$ $\Delta G^{\ominus} < 0, \Delta S^{\ominus} > 0$
Calculate the partial pressure of carbon monoxide from the following data:
$CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}, K_p = 8 \times 10^{-2}$
$CO_{2(g)} + C_{(s)} \rightleftharpoons 2CO_{(g)}, K_p = 2$
$CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}, K_p = 8 \times 10^{-2}$
$CO_{2(g)} + C_{(s)} \rightleftharpoons 2CO_{(g)}, K_p = 2$
Difficult
View SolutionFor the equilibrium reaction $A + B \rightleftharpoons C + D$,if we start with equal concentrations of $A$ and $B$,at equilibrium,the concentration of $C$ is $2$ times that of $A$. Find the value of $K_c$.
Medium
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