229321 For the equilibrium, $2 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{OH}^{-}$, the value of $\Delta \mathbf{G}^{\mathbf{0}}$ at $298 \mathrm{~K}$ is approximately

229322 Equivalent amounts of $\mathrm{H}_{2}$ and $\mathrm{I}_{2}$ are heated in a closed vessel till equilibrium is obtained. If $80 \%$ of the hydrogen can be converted to HI, the $K_{c}$ at the temperature is

229323 For the reaction,
$\mathrm{Fe}_{2} \mathrm{~N}(\mathrm{~s})+\frac{\mathbf{3}}{\mathbf{2}} \mathrm{H}_{2}(\mathrm{~g})\rightleftharpoons \mathrm{Fe}(\mathrm{s})+\mathrm{NH}_{3}(\mathrm{~g})$

229324 In a chemical reaction,
$A+2 B \stackrel{k}{\rightleftharpoons} 2 C+D$, the initial concentration of $B$ was 1.5 times of the concentration of $A$, but the equilibrium concentrations of $A$ and $B$ were found to be equal. The equilibrium constant $(K)$ for the chemical reaction is

229325 The value of $K_{c}$ for the reaction :
$\mathrm{A}+3 \mathrm{~B} \rightleftharpoons \quad 2 \mathrm{C}$ at $400^{\circ} \mathrm{C}$ is 0.5 Calculate the value of $\mathbf{K}_{\mathbf{p}}$
#[Qdiff: Hard, QCat: Numerical Based, examname: VITEEE-2018, VITEEE-2019, $\begin{aligned}, & \text { Ans. A, Exp: Given, \end{aligned}$, } \\, & \mathrm{K}_{\mathrm{c}}=0.5 \\, & \mathrm{~T}=400^{\circ} \mathrm{C}=400+273=673 \mathrm{~K} \\, & \mathrm{~K}_{\mathrm{p}}=? \\, & \mathrm{~A}+3 \mathrm{~B} \rightleftharpoons \quad 2 \mathrm{C} \\, & \Delta \mathrm{n}=2-4=-2 \\, & \mathrm{~K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{\mathrm{\Delta n}} \\, & \mathrm{K}_{\mathrm{p}}=0.5 \times(0.082 \times 673)^{-2} \\, & \mathrm{~K}_{\mathrm{p}}=1.64 \times 10^{-4} \text { atm. }, 434. For the reaction:, $2 \mathrm{BaO}_{2}(\mathrm{~s}) \rightleftharpoons \quad 2 \mathrm{BaO}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g})$;, $\Delta H=+v e$. In equilibrium condition, pressure of $\mathrm{O}_{2}$ is dependent on,

229321 For the equilibrium, $2 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{OH}^{-}$, the value of $\Delta \mathbf{G}^{\mathbf{0}}$ at $298 \mathrm{~K}$ is approximately

229322 Equivalent amounts of $\mathrm{H}_{2}$ and $\mathrm{I}_{2}$ are heated in a closed vessel till equilibrium is obtained. If $80 \%$ of the hydrogen can be converted to HI, the $K_{c}$ at the temperature is

229323 For the reaction,
$\mathrm{Fe}_{2} \mathrm{~N}(\mathrm{~s})+\frac{\mathbf{3}}{\mathbf{2}} \mathrm{H}_{2}(\mathrm{~g})\rightleftharpoons \mathrm{Fe}(\mathrm{s})+\mathrm{NH}_{3}(\mathrm{~g})$

229324 In a chemical reaction,
$A+2 B \stackrel{k}{\rightleftharpoons} 2 C+D$, the initial concentration of $B$ was 1.5 times of the concentration of $A$, but the equilibrium concentrations of $A$ and $B$ were found to be equal. The equilibrium constant $(K)$ for the chemical reaction is

229325 The value of $K_{c}$ for the reaction :
$\mathrm{A}+3 \mathrm{~B} \rightleftharpoons \quad 2 \mathrm{C}$ at $400^{\circ} \mathrm{C}$ is 0.5 Calculate the value of $\mathbf{K}_{\mathbf{p}}$
#[Qdiff: Hard, QCat: Numerical Based, examname: VITEEE-2018, VITEEE-2019, $\begin{aligned}, & \text { Ans. A, Exp: Given, \end{aligned}$, } \\, & \mathrm{K}_{\mathrm{c}}=0.5 \\, & \mathrm{~T}=400^{\circ} \mathrm{C}=400+273=673 \mathrm{~K} \\, & \mathrm{~K}_{\mathrm{p}}=? \\, & \mathrm{~A}+3 \mathrm{~B} \rightleftharpoons \quad 2 \mathrm{C} \\, & \Delta \mathrm{n}=2-4=-2 \\, & \mathrm{~K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{\mathrm{\Delta n}} \\, & \mathrm{K}_{\mathrm{p}}=0.5 \times(0.082 \times 673)^{-2} \\, & \mathrm{~K}_{\mathrm{p}}=1.64 \times 10^{-4} \text { atm. }, 434. For the reaction:, $2 \mathrm{BaO}_{2}(\mathrm{~s}) \rightleftharpoons \quad 2 \mathrm{BaO}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g})$;, $\Delta H=+v e$. In equilibrium condition, pressure of $\mathrm{O}_{2}$ is dependent on,

229321 For the equilibrium, $2 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{OH}^{-}$, the value of $\Delta \mathbf{G}^{\mathbf{0}}$ at $298 \mathrm{~K}$ is approximately

229322 Equivalent amounts of $\mathrm{H}_{2}$ and $\mathrm{I}_{2}$ are heated in a closed vessel till equilibrium is obtained. If $80 \%$ of the hydrogen can be converted to HI, the $K_{c}$ at the temperature is

229323 For the reaction,
$\mathrm{Fe}_{2} \mathrm{~N}(\mathrm{~s})+\frac{\mathbf{3}}{\mathbf{2}} \mathrm{H}_{2}(\mathrm{~g})\rightleftharpoons \mathrm{Fe}(\mathrm{s})+\mathrm{NH}_{3}(\mathrm{~g})$

229324 In a chemical reaction,
$A+2 B \stackrel{k}{\rightleftharpoons} 2 C+D$, the initial concentration of $B$ was 1.5 times of the concentration of $A$, but the equilibrium concentrations of $A$ and $B$ were found to be equal. The equilibrium constant $(K)$ for the chemical reaction is

229325 The value of $K_{c}$ for the reaction :
$\mathrm{A}+3 \mathrm{~B} \rightleftharpoons \quad 2 \mathrm{C}$ at $400^{\circ} \mathrm{C}$ is 0.5 Calculate the value of $\mathbf{K}_{\mathbf{p}}$
#[Qdiff: Hard, QCat: Numerical Based, examname: VITEEE-2018, VITEEE-2019, $\begin{aligned}, & \text { Ans. A, Exp: Given, \end{aligned}$, } \\, & \mathrm{K}_{\mathrm{c}}=0.5 \\, & \mathrm{~T}=400^{\circ} \mathrm{C}=400+273=673 \mathrm{~K} \\, & \mathrm{~K}_{\mathrm{p}}=? \\, & \mathrm{~A}+3 \mathrm{~B} \rightleftharpoons \quad 2 \mathrm{C} \\, & \Delta \mathrm{n}=2-4=-2 \\, & \mathrm{~K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{\mathrm{\Delta n}} \\, & \mathrm{K}_{\mathrm{p}}=0.5 \times(0.082 \times 673)^{-2} \\, & \mathrm{~K}_{\mathrm{p}}=1.64 \times 10^{-4} \text { atm. }, 434. For the reaction:, $2 \mathrm{BaO}_{2}(\mathrm{~s}) \rightleftharpoons \quad 2 \mathrm{BaO}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g})$;, $\Delta H=+v e$. In equilibrium condition, pressure of $\mathrm{O}_{2}$ is dependent on,

229321 For the equilibrium, $2 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{OH}^{-}$, the value of $\Delta \mathbf{G}^{\mathbf{0}}$ at $298 \mathrm{~K}$ is approximately

229322 Equivalent amounts of $\mathrm{H}_{2}$ and $\mathrm{I}_{2}$ are heated in a closed vessel till equilibrium is obtained. If $80 \%$ of the hydrogen can be converted to HI, the $K_{c}$ at the temperature is

229323 For the reaction,
$\mathrm{Fe}_{2} \mathrm{~N}(\mathrm{~s})+\frac{\mathbf{3}}{\mathbf{2}} \mathrm{H}_{2}(\mathrm{~g})\rightleftharpoons \mathrm{Fe}(\mathrm{s})+\mathrm{NH}_{3}(\mathrm{~g})$

229324 In a chemical reaction,
$A+2 B \stackrel{k}{\rightleftharpoons} 2 C+D$, the initial concentration of $B$ was 1.5 times of the concentration of $A$, but the equilibrium concentrations of $A$ and $B$ were found to be equal. The equilibrium constant $(K)$ for the chemical reaction is

229325 The value of $K_{c}$ for the reaction :
$\mathrm{A}+3 \mathrm{~B} \rightleftharpoons \quad 2 \mathrm{C}$ at $400^{\circ} \mathrm{C}$ is 0.5 Calculate the value of $\mathbf{K}_{\mathbf{p}}$
#[Qdiff: Hard, QCat: Numerical Based, examname: VITEEE-2018, VITEEE-2019, $\begin{aligned}, & \text { Ans. A, Exp: Given, \end{aligned}$, } \\, & \mathrm{K}_{\mathrm{c}}=0.5 \\, & \mathrm{~T}=400^{\circ} \mathrm{C}=400+273=673 \mathrm{~K} \\, & \mathrm{~K}_{\mathrm{p}}=? \\, & \mathrm{~A}+3 \mathrm{~B} \rightleftharpoons \quad 2 \mathrm{C} \\, & \Delta \mathrm{n}=2-4=-2 \\, & \mathrm{~K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{\mathrm{\Delta n}} \\, & \mathrm{K}_{\mathrm{p}}=0.5 \times(0.082 \times 673)^{-2} \\, & \mathrm{~K}_{\mathrm{p}}=1.64 \times 10^{-4} \text { atm. }, 434. For the reaction:, $2 \mathrm{BaO}_{2}(\mathrm{~s}) \rightleftharpoons \quad 2 \mathrm{BaO}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g})$;, $\Delta H=+v e$. In equilibrium condition, pressure of $\mathrm{O}_{2}$ is dependent on,

229321 For the equilibrium, $2 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{OH}^{-}$, the value of $\Delta \mathbf{G}^{\mathbf{0}}$ at $298 \mathrm{~K}$ is approximately

229322 Equivalent amounts of $\mathrm{H}_{2}$ and $\mathrm{I}_{2}$ are heated in a closed vessel till equilibrium is obtained. If $80 \%$ of the hydrogen can be converted to HI, the $K_{c}$ at the temperature is

229323 For the reaction,
$\mathrm{Fe}_{2} \mathrm{~N}(\mathrm{~s})+\frac{\mathbf{3}}{\mathbf{2}} \mathrm{H}_{2}(\mathrm{~g})\rightleftharpoons \mathrm{Fe}(\mathrm{s})+\mathrm{NH}_{3}(\mathrm{~g})$

229324 In a chemical reaction,
$A+2 B \stackrel{k}{\rightleftharpoons} 2 C+D$, the initial concentration of $B$ was 1.5 times of the concentration of $A$, but the equilibrium concentrations of $A$ and $B$ were found to be equal. The equilibrium constant $(K)$ for the chemical reaction is

229325 The value of $K_{c}$ for the reaction :
$\mathrm{A}+3 \mathrm{~B} \rightleftharpoons \quad 2 \mathrm{C}$ at $400^{\circ} \mathrm{C}$ is 0.5 Calculate the value of $\mathbf{K}_{\mathbf{p}}$
#[Qdiff: Hard, QCat: Numerical Based, examname: VITEEE-2018, VITEEE-2019, $\begin{aligned}, & \text { Ans. A, Exp: Given, \end{aligned}$, } \\, & \mathrm{K}_{\mathrm{c}}=0.5 \\, & \mathrm{~T}=400^{\circ} \mathrm{C}=400+273=673 \mathrm{~K} \\, & \mathrm{~K}_{\mathrm{p}}=? \\, & \mathrm{~A}+3 \mathrm{~B} \rightleftharpoons \quad 2 \mathrm{C} \\, & \Delta \mathrm{n}=2-4=-2 \\, & \mathrm{~K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{\mathrm{\Delta n}} \\, & \mathrm{K}_{\mathrm{p}}=0.5 \times(0.082 \times 673)^{-2} \\, & \mathrm{~K}_{\mathrm{p}}=1.64 \times 10^{-4} \text { atm. }, 434. For the reaction:, $2 \mathrm{BaO}_{2}(\mathrm{~s}) \rightleftharpoons \quad 2 \mathrm{BaO}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g})$;, $\Delta H=+v e$. In equilibrium condition, pressure of $\mathrm{O}_{2}$ is dependent on,