229333
Which of the following is a wrong statement about equilibrium state?
1 Rate of forward reaction= Rate of backward reaction
2 Equilibrium is dynamic
3 Catalysts increase value of equilibrium constant
4 Free energy change is zero
Explanation:
Chemical equilibrium is dynamic in nature which in other words means that reactions continue to occur in both forward as well as backward direction with the same speed. Equilibrium constant has definite value for every chemical reaction at a given temperature. It is independent of concentration and catalyst. Free energy change : $\Delta \mathrm{G}=\Delta \mathrm{G}^{\mathrm{o}}+2.303 \mathrm{RT} \log \mathrm{Q}$ So, option (a), (b), (d) follows equilibrium state but (c) not follows. Hence, option (c) is correct.
CG PET -2009
Chemical Equilibrium
229335
The standard emf of a cell, involving one electron change is found to be $0.591 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction is $(F=$ $96500 \mathrm{Cmol}^{-1}$ )
1 $1.0 \times 10^{1}$
2 $1.0 \times 10^{5}$
3 $1.0 \times 10^{10}$
4 $1.0 \times 10^{30}$
Explanation:
Relation between $\mathrm{K}_{\mathrm{eq}}$ and $\mathrm{E}_{\text {cell }}$ us $\mathrm{E}_{\text {cell }}^{\circ}=\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log \mathrm{K}_{\mathrm{eq}}$ $\mathrm{E}_{\text {cell }}^{\circ}=\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log \mathrm{K}_{\mathrm{eq}}($ at $298 \mathrm{~K})$ $\mathrm{E}_{\text {cell }}^{\circ}=\frac{0.0591}{\mathrm{n}} \log \mathrm{K}_{\mathrm{eq}}$ $0.0591=\frac{0.0591}{\mathrm{n}} \log \mathrm{K}_{\mathrm{eq}}$ $\log \mathrm{K}_{\mathrm{eq}}=10=\mathrm{K}_{\mathrm{eq}}=\operatorname{antilog} 10$ $\mathrm{K}_{\mathrm{eq}}=1 \times 10^{10}$
UPTU/UPSEE-2008
Chemical Equilibrium
229336
The equilibrium constant of the reaction $\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})$ [Given : $\mathrm{E}^{\mathrm{o}}=0.46 \mathrm{~V}$ ] will be
1 $4 \times 10^{15}$
2 $4 \times 10^{14}$
3 $14 \times 10^{15}$
4 $15.6 \times 10^{5}$
Explanation:
The equilibrium constant for the give reaction can be given as, [Given : $\mathrm{E}^{\mathrm{o}}=0.46 \mathrm{~V}$ ] $\begin{aligned} & \mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s}) \\ & \mathrm{E}^{\circ}=\frac{0.0591}{\mathrm{n}} \log \mathrm{K} \end{aligned}$ Where, $\mathrm{K}=$ Equilibrium constant, $\mathrm{n}=$ Number of electrons. On putting the values in eqn, ....(i) we get $0.46=\frac{0.0591}{\mathrm{n}} \log \mathrm{K}$ $\log \mathrm{K}=\frac{0.46 \times 2}{0.0591}$ $\therefore \quad \mathrm{K}=$ Antilog 15.6 $=3.688 \times 10^{15}$ or $=4 \times 10^{15}$
AMU-2007
Chemical Equilibrium
229337
The standard emf of a cell, involving one electron change is found to be $0.591 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction is $(F=$ $96500 \mathrm{C} \mathrm{mol}^{-1}$ )
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Chemical Equilibrium
229333
Which of the following is a wrong statement about equilibrium state?
1 Rate of forward reaction= Rate of backward reaction
2 Equilibrium is dynamic
3 Catalysts increase value of equilibrium constant
4 Free energy change is zero
Explanation:
Chemical equilibrium is dynamic in nature which in other words means that reactions continue to occur in both forward as well as backward direction with the same speed. Equilibrium constant has definite value for every chemical reaction at a given temperature. It is independent of concentration and catalyst. Free energy change : $\Delta \mathrm{G}=\Delta \mathrm{G}^{\mathrm{o}}+2.303 \mathrm{RT} \log \mathrm{Q}$ So, option (a), (b), (d) follows equilibrium state but (c) not follows. Hence, option (c) is correct.
CG PET -2009
Chemical Equilibrium
229335
The standard emf of a cell, involving one electron change is found to be $0.591 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction is $(F=$ $96500 \mathrm{Cmol}^{-1}$ )
1 $1.0 \times 10^{1}$
2 $1.0 \times 10^{5}$
3 $1.0 \times 10^{10}$
4 $1.0 \times 10^{30}$
Explanation:
Relation between $\mathrm{K}_{\mathrm{eq}}$ and $\mathrm{E}_{\text {cell }}$ us $\mathrm{E}_{\text {cell }}^{\circ}=\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log \mathrm{K}_{\mathrm{eq}}$ $\mathrm{E}_{\text {cell }}^{\circ}=\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log \mathrm{K}_{\mathrm{eq}}($ at $298 \mathrm{~K})$ $\mathrm{E}_{\text {cell }}^{\circ}=\frac{0.0591}{\mathrm{n}} \log \mathrm{K}_{\mathrm{eq}}$ $0.0591=\frac{0.0591}{\mathrm{n}} \log \mathrm{K}_{\mathrm{eq}}$ $\log \mathrm{K}_{\mathrm{eq}}=10=\mathrm{K}_{\mathrm{eq}}=\operatorname{antilog} 10$ $\mathrm{K}_{\mathrm{eq}}=1 \times 10^{10}$
UPTU/UPSEE-2008
Chemical Equilibrium
229336
The equilibrium constant of the reaction $\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})$ [Given : $\mathrm{E}^{\mathrm{o}}=0.46 \mathrm{~V}$ ] will be
1 $4 \times 10^{15}$
2 $4 \times 10^{14}$
3 $14 \times 10^{15}$
4 $15.6 \times 10^{5}$
Explanation:
The equilibrium constant for the give reaction can be given as, [Given : $\mathrm{E}^{\mathrm{o}}=0.46 \mathrm{~V}$ ] $\begin{aligned} & \mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s}) \\ & \mathrm{E}^{\circ}=\frac{0.0591}{\mathrm{n}} \log \mathrm{K} \end{aligned}$ Where, $\mathrm{K}=$ Equilibrium constant, $\mathrm{n}=$ Number of electrons. On putting the values in eqn, ....(i) we get $0.46=\frac{0.0591}{\mathrm{n}} \log \mathrm{K}$ $\log \mathrm{K}=\frac{0.46 \times 2}{0.0591}$ $\therefore \quad \mathrm{K}=$ Antilog 15.6 $=3.688 \times 10^{15}$ or $=4 \times 10^{15}$
AMU-2007
Chemical Equilibrium
229337
The standard emf of a cell, involving one electron change is found to be $0.591 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction is $(F=$ $96500 \mathrm{C} \mathrm{mol}^{-1}$ )
229333
Which of the following is a wrong statement about equilibrium state?
1 Rate of forward reaction= Rate of backward reaction
2 Equilibrium is dynamic
3 Catalysts increase value of equilibrium constant
4 Free energy change is zero
Explanation:
Chemical equilibrium is dynamic in nature which in other words means that reactions continue to occur in both forward as well as backward direction with the same speed. Equilibrium constant has definite value for every chemical reaction at a given temperature. It is independent of concentration and catalyst. Free energy change : $\Delta \mathrm{G}=\Delta \mathrm{G}^{\mathrm{o}}+2.303 \mathrm{RT} \log \mathrm{Q}$ So, option (a), (b), (d) follows equilibrium state but (c) not follows. Hence, option (c) is correct.
CG PET -2009
Chemical Equilibrium
229335
The standard emf of a cell, involving one electron change is found to be $0.591 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction is $(F=$ $96500 \mathrm{Cmol}^{-1}$ )
1 $1.0 \times 10^{1}$
2 $1.0 \times 10^{5}$
3 $1.0 \times 10^{10}$
4 $1.0 \times 10^{30}$
Explanation:
Relation between $\mathrm{K}_{\mathrm{eq}}$ and $\mathrm{E}_{\text {cell }}$ us $\mathrm{E}_{\text {cell }}^{\circ}=\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log \mathrm{K}_{\mathrm{eq}}$ $\mathrm{E}_{\text {cell }}^{\circ}=\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log \mathrm{K}_{\mathrm{eq}}($ at $298 \mathrm{~K})$ $\mathrm{E}_{\text {cell }}^{\circ}=\frac{0.0591}{\mathrm{n}} \log \mathrm{K}_{\mathrm{eq}}$ $0.0591=\frac{0.0591}{\mathrm{n}} \log \mathrm{K}_{\mathrm{eq}}$ $\log \mathrm{K}_{\mathrm{eq}}=10=\mathrm{K}_{\mathrm{eq}}=\operatorname{antilog} 10$ $\mathrm{K}_{\mathrm{eq}}=1 \times 10^{10}$
UPTU/UPSEE-2008
Chemical Equilibrium
229336
The equilibrium constant of the reaction $\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})$ [Given : $\mathrm{E}^{\mathrm{o}}=0.46 \mathrm{~V}$ ] will be
1 $4 \times 10^{15}$
2 $4 \times 10^{14}$
3 $14 \times 10^{15}$
4 $15.6 \times 10^{5}$
Explanation:
The equilibrium constant for the give reaction can be given as, [Given : $\mathrm{E}^{\mathrm{o}}=0.46 \mathrm{~V}$ ] $\begin{aligned} & \mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s}) \\ & \mathrm{E}^{\circ}=\frac{0.0591}{\mathrm{n}} \log \mathrm{K} \end{aligned}$ Where, $\mathrm{K}=$ Equilibrium constant, $\mathrm{n}=$ Number of electrons. On putting the values in eqn, ....(i) we get $0.46=\frac{0.0591}{\mathrm{n}} \log \mathrm{K}$ $\log \mathrm{K}=\frac{0.46 \times 2}{0.0591}$ $\therefore \quad \mathrm{K}=$ Antilog 15.6 $=3.688 \times 10^{15}$ or $=4 \times 10^{15}$
AMU-2007
Chemical Equilibrium
229337
The standard emf of a cell, involving one electron change is found to be $0.591 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction is $(F=$ $96500 \mathrm{C} \mathrm{mol}^{-1}$ )
229333
Which of the following is a wrong statement about equilibrium state?
1 Rate of forward reaction= Rate of backward reaction
2 Equilibrium is dynamic
3 Catalysts increase value of equilibrium constant
4 Free energy change is zero
Explanation:
Chemical equilibrium is dynamic in nature which in other words means that reactions continue to occur in both forward as well as backward direction with the same speed. Equilibrium constant has definite value for every chemical reaction at a given temperature. It is independent of concentration and catalyst. Free energy change : $\Delta \mathrm{G}=\Delta \mathrm{G}^{\mathrm{o}}+2.303 \mathrm{RT} \log \mathrm{Q}$ So, option (a), (b), (d) follows equilibrium state but (c) not follows. Hence, option (c) is correct.
CG PET -2009
Chemical Equilibrium
229335
The standard emf of a cell, involving one electron change is found to be $0.591 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction is $(F=$ $96500 \mathrm{Cmol}^{-1}$ )
1 $1.0 \times 10^{1}$
2 $1.0 \times 10^{5}$
3 $1.0 \times 10^{10}$
4 $1.0 \times 10^{30}$
Explanation:
Relation between $\mathrm{K}_{\mathrm{eq}}$ and $\mathrm{E}_{\text {cell }}$ us $\mathrm{E}_{\text {cell }}^{\circ}=\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log \mathrm{K}_{\mathrm{eq}}$ $\mathrm{E}_{\text {cell }}^{\circ}=\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log \mathrm{K}_{\mathrm{eq}}($ at $298 \mathrm{~K})$ $\mathrm{E}_{\text {cell }}^{\circ}=\frac{0.0591}{\mathrm{n}} \log \mathrm{K}_{\mathrm{eq}}$ $0.0591=\frac{0.0591}{\mathrm{n}} \log \mathrm{K}_{\mathrm{eq}}$ $\log \mathrm{K}_{\mathrm{eq}}=10=\mathrm{K}_{\mathrm{eq}}=\operatorname{antilog} 10$ $\mathrm{K}_{\mathrm{eq}}=1 \times 10^{10}$
UPTU/UPSEE-2008
Chemical Equilibrium
229336
The equilibrium constant of the reaction $\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})$ [Given : $\mathrm{E}^{\mathrm{o}}=0.46 \mathrm{~V}$ ] will be
1 $4 \times 10^{15}$
2 $4 \times 10^{14}$
3 $14 \times 10^{15}$
4 $15.6 \times 10^{5}$
Explanation:
The equilibrium constant for the give reaction can be given as, [Given : $\mathrm{E}^{\mathrm{o}}=0.46 \mathrm{~V}$ ] $\begin{aligned} & \mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s}) \\ & \mathrm{E}^{\circ}=\frac{0.0591}{\mathrm{n}} \log \mathrm{K} \end{aligned}$ Where, $\mathrm{K}=$ Equilibrium constant, $\mathrm{n}=$ Number of electrons. On putting the values in eqn, ....(i) we get $0.46=\frac{0.0591}{\mathrm{n}} \log \mathrm{K}$ $\log \mathrm{K}=\frac{0.46 \times 2}{0.0591}$ $\therefore \quad \mathrm{K}=$ Antilog 15.6 $=3.688 \times 10^{15}$ or $=4 \times 10^{15}$
AMU-2007
Chemical Equilibrium
229337
The standard emf of a cell, involving one electron change is found to be $0.591 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction is $(F=$ $96500 \mathrm{C} \mathrm{mol}^{-1}$ )