229308
In which of the following reactions, the concentration of product is higher than the concentration of reactant at equilibrium? ( $K_{c}=$ equilibrium constant)
For a reaction $\mathrm{K}_{\mathrm{c}}=\frac{[\text { Product }]}{[\text { reactant }]}$ Hence, If $\mathrm{K}_{\mathrm{c}}>1$, then, [Product] > [Reactant]. As per option, (a) $\mathrm{K}_{\mathrm{c}}=0.001$ which is $<1$ (c) $\mathrm{K}_{\mathrm{c}}=0.005$ which is $<1$ (d) $\mathrm{K}_{\mathrm{c}}=0.01$ which is $<1$ So, in option (b) $\mathrm{K}_{\mathrm{c}}=10$ which is $>1$. Hence, (b) is correct answer.
A.P.EAMCET 2004
Chemical Equilibrium
229309
For reaction, $\mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{5}(\mathrm{~g})$ the value of $K_{c}$ at $250^{\circ} \mathrm{C}$ is 26 . At the same temperature, the value of $K_{p}$ is:
1 0.46
2 0.61
3 0.95
4 0.73
Explanation:
Given data: $\mathrm{K}_{\mathrm{c}}=26, \mathrm{~T}=250^{\circ} \mathrm{C}=250+273$ $\mathrm{K}_{\mathrm{p}}=? \quad=523 \mathrm{~K}$ For the reaction, $\mathrm{PCl}_{3}+\mathrm{Cl}_{2} \rightleftharpoons \mathrm{PCl}_{5}$ $\therefore \quad \mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{\Delta \mathrm{n}_{\mathrm{g}}}$ $\Delta \mathrm{n}_{\mathrm{g}}=1-2=-1$ $\mathrm{R}=0.0821, \mathrm{~T}=250+273.15=523.15$ So, $\quad \mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{-1}$ $\begin{aligned} \mathrm{K}_{\mathrm{p}} & =\frac{26}{0.0821 \times 523.15} \\ \mathrm{~K}_{\mathrm{p}} & =0.61 \end{aligned}$
JCECE - 2003
Chemical Equilibrium
229310
Equilibrium constant $K_{p}$ of following reaction $\mathrm{MgCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})$
The given reaction is - $\mathrm{MgCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})$ $\therefore \quad \mathrm{K}_{\mathrm{p}}=\frac{\mathrm{p}_{\mathrm{CO}_{2}} \cdot \mathrm{p}_{\mathrm{MgO}}}{\mathrm{p}_{\mathrm{MgCO}_{3}}}$ But $\mathrm{MgCO}_{3}$ and $\mathrm{MgO}$ are pure solids So, $\quad \mathrm{K}_{\mathrm{p}}=\mathrm{p}_{\mathrm{CO}_{2}}$
NEET-2000
Chemical Equilibrium
229311
4.5 moles, each of hydrogen and iodine was heated in a sealed $10 \mathrm{~L}$ vessel. At equilibrium, 3 moles of $\mathrm{HI}$ were found. The equilibrium constant for $\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{HI}(\mathrm{g})$, is
229308
In which of the following reactions, the concentration of product is higher than the concentration of reactant at equilibrium? ( $K_{c}=$ equilibrium constant)
For a reaction $\mathrm{K}_{\mathrm{c}}=\frac{[\text { Product }]}{[\text { reactant }]}$ Hence, If $\mathrm{K}_{\mathrm{c}}>1$, then, [Product] > [Reactant]. As per option, (a) $\mathrm{K}_{\mathrm{c}}=0.001$ which is $<1$ (c) $\mathrm{K}_{\mathrm{c}}=0.005$ which is $<1$ (d) $\mathrm{K}_{\mathrm{c}}=0.01$ which is $<1$ So, in option (b) $\mathrm{K}_{\mathrm{c}}=10$ which is $>1$. Hence, (b) is correct answer.
A.P.EAMCET 2004
Chemical Equilibrium
229309
For reaction, $\mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{5}(\mathrm{~g})$ the value of $K_{c}$ at $250^{\circ} \mathrm{C}$ is 26 . At the same temperature, the value of $K_{p}$ is:
1 0.46
2 0.61
3 0.95
4 0.73
Explanation:
Given data: $\mathrm{K}_{\mathrm{c}}=26, \mathrm{~T}=250^{\circ} \mathrm{C}=250+273$ $\mathrm{K}_{\mathrm{p}}=? \quad=523 \mathrm{~K}$ For the reaction, $\mathrm{PCl}_{3}+\mathrm{Cl}_{2} \rightleftharpoons \mathrm{PCl}_{5}$ $\therefore \quad \mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{\Delta \mathrm{n}_{\mathrm{g}}}$ $\Delta \mathrm{n}_{\mathrm{g}}=1-2=-1$ $\mathrm{R}=0.0821, \mathrm{~T}=250+273.15=523.15$ So, $\quad \mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{-1}$ $\begin{aligned} \mathrm{K}_{\mathrm{p}} & =\frac{26}{0.0821 \times 523.15} \\ \mathrm{~K}_{\mathrm{p}} & =0.61 \end{aligned}$
JCECE - 2003
Chemical Equilibrium
229310
Equilibrium constant $K_{p}$ of following reaction $\mathrm{MgCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})$
The given reaction is - $\mathrm{MgCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})$ $\therefore \quad \mathrm{K}_{\mathrm{p}}=\frac{\mathrm{p}_{\mathrm{CO}_{2}} \cdot \mathrm{p}_{\mathrm{MgO}}}{\mathrm{p}_{\mathrm{MgCO}_{3}}}$ But $\mathrm{MgCO}_{3}$ and $\mathrm{MgO}$ are pure solids So, $\quad \mathrm{K}_{\mathrm{p}}=\mathrm{p}_{\mathrm{CO}_{2}}$
NEET-2000
Chemical Equilibrium
229311
4.5 moles, each of hydrogen and iodine was heated in a sealed $10 \mathrm{~L}$ vessel. At equilibrium, 3 moles of $\mathrm{HI}$ were found. The equilibrium constant for $\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{HI}(\mathrm{g})$, is
229308
In which of the following reactions, the concentration of product is higher than the concentration of reactant at equilibrium? ( $K_{c}=$ equilibrium constant)
For a reaction $\mathrm{K}_{\mathrm{c}}=\frac{[\text { Product }]}{[\text { reactant }]}$ Hence, If $\mathrm{K}_{\mathrm{c}}>1$, then, [Product] > [Reactant]. As per option, (a) $\mathrm{K}_{\mathrm{c}}=0.001$ which is $<1$ (c) $\mathrm{K}_{\mathrm{c}}=0.005$ which is $<1$ (d) $\mathrm{K}_{\mathrm{c}}=0.01$ which is $<1$ So, in option (b) $\mathrm{K}_{\mathrm{c}}=10$ which is $>1$. Hence, (b) is correct answer.
A.P.EAMCET 2004
Chemical Equilibrium
229309
For reaction, $\mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{5}(\mathrm{~g})$ the value of $K_{c}$ at $250^{\circ} \mathrm{C}$ is 26 . At the same temperature, the value of $K_{p}$ is:
1 0.46
2 0.61
3 0.95
4 0.73
Explanation:
Given data: $\mathrm{K}_{\mathrm{c}}=26, \mathrm{~T}=250^{\circ} \mathrm{C}=250+273$ $\mathrm{K}_{\mathrm{p}}=? \quad=523 \mathrm{~K}$ For the reaction, $\mathrm{PCl}_{3}+\mathrm{Cl}_{2} \rightleftharpoons \mathrm{PCl}_{5}$ $\therefore \quad \mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{\Delta \mathrm{n}_{\mathrm{g}}}$ $\Delta \mathrm{n}_{\mathrm{g}}=1-2=-1$ $\mathrm{R}=0.0821, \mathrm{~T}=250+273.15=523.15$ So, $\quad \mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{-1}$ $\begin{aligned} \mathrm{K}_{\mathrm{p}} & =\frac{26}{0.0821 \times 523.15} \\ \mathrm{~K}_{\mathrm{p}} & =0.61 \end{aligned}$
JCECE - 2003
Chemical Equilibrium
229310
Equilibrium constant $K_{p}$ of following reaction $\mathrm{MgCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})$
The given reaction is - $\mathrm{MgCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})$ $\therefore \quad \mathrm{K}_{\mathrm{p}}=\frac{\mathrm{p}_{\mathrm{CO}_{2}} \cdot \mathrm{p}_{\mathrm{MgO}}}{\mathrm{p}_{\mathrm{MgCO}_{3}}}$ But $\mathrm{MgCO}_{3}$ and $\mathrm{MgO}$ are pure solids So, $\quad \mathrm{K}_{\mathrm{p}}=\mathrm{p}_{\mathrm{CO}_{2}}$
NEET-2000
Chemical Equilibrium
229311
4.5 moles, each of hydrogen and iodine was heated in a sealed $10 \mathrm{~L}$ vessel. At equilibrium, 3 moles of $\mathrm{HI}$ were found. The equilibrium constant for $\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{HI}(\mathrm{g})$, is
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Chemical Equilibrium
229308
In which of the following reactions, the concentration of product is higher than the concentration of reactant at equilibrium? ( $K_{c}=$ equilibrium constant)
For a reaction $\mathrm{K}_{\mathrm{c}}=\frac{[\text { Product }]}{[\text { reactant }]}$ Hence, If $\mathrm{K}_{\mathrm{c}}>1$, then, [Product] > [Reactant]. As per option, (a) $\mathrm{K}_{\mathrm{c}}=0.001$ which is $<1$ (c) $\mathrm{K}_{\mathrm{c}}=0.005$ which is $<1$ (d) $\mathrm{K}_{\mathrm{c}}=0.01$ which is $<1$ So, in option (b) $\mathrm{K}_{\mathrm{c}}=10$ which is $>1$. Hence, (b) is correct answer.
A.P.EAMCET 2004
Chemical Equilibrium
229309
For reaction, $\mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{5}(\mathrm{~g})$ the value of $K_{c}$ at $250^{\circ} \mathrm{C}$ is 26 . At the same temperature, the value of $K_{p}$ is:
1 0.46
2 0.61
3 0.95
4 0.73
Explanation:
Given data: $\mathrm{K}_{\mathrm{c}}=26, \mathrm{~T}=250^{\circ} \mathrm{C}=250+273$ $\mathrm{K}_{\mathrm{p}}=? \quad=523 \mathrm{~K}$ For the reaction, $\mathrm{PCl}_{3}+\mathrm{Cl}_{2} \rightleftharpoons \mathrm{PCl}_{5}$ $\therefore \quad \mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{\Delta \mathrm{n}_{\mathrm{g}}}$ $\Delta \mathrm{n}_{\mathrm{g}}=1-2=-1$ $\mathrm{R}=0.0821, \mathrm{~T}=250+273.15=523.15$ So, $\quad \mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{-1}$ $\begin{aligned} \mathrm{K}_{\mathrm{p}} & =\frac{26}{0.0821 \times 523.15} \\ \mathrm{~K}_{\mathrm{p}} & =0.61 \end{aligned}$
JCECE - 2003
Chemical Equilibrium
229310
Equilibrium constant $K_{p}$ of following reaction $\mathrm{MgCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})$
The given reaction is - $\mathrm{MgCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})$ $\therefore \quad \mathrm{K}_{\mathrm{p}}=\frac{\mathrm{p}_{\mathrm{CO}_{2}} \cdot \mathrm{p}_{\mathrm{MgO}}}{\mathrm{p}_{\mathrm{MgCO}_{3}}}$ But $\mathrm{MgCO}_{3}$ and $\mathrm{MgO}$ are pure solids So, $\quad \mathrm{K}_{\mathrm{p}}=\mathrm{p}_{\mathrm{CO}_{2}}$
NEET-2000
Chemical Equilibrium
229311
4.5 moles, each of hydrogen and iodine was heated in a sealed $10 \mathrm{~L}$ vessel. At equilibrium, 3 moles of $\mathrm{HI}$ were found. The equilibrium constant for $\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{HI}(\mathrm{g})$, is
