276091 The resistance of $0.01 \mathrm{~N}$ solution of an electrolyte was found to be $220 \mathrm{ohm}$ at $298 \mathrm{~K}$ using a conductivity cell with a cell constant of $0.88 \mathrm{~cm}^{-1}$. The value of equivalent conductance of solution is -

276092 The EMF of the cell,
$\mathrm{Mg}\left \vert\mathrm{Mg}^{2+}(0.01 \mathrm{M})\right \vert\left \vert\mathrm{Sn}^{2+}(0.1 \mathrm{M})\right \vert \mathrm{Sn}$ at $298 \mathrm{~K}$ is
$\left(\mathrm{E}_{\mathrm{Mg}^{2+} / \mathrm{Mg}}^{\circ}=-2.34 \mathrm{~V}, \mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{\circ}=-0.14 \mathrm{~V}\right)$

276093 In the reversible reaction.
$2 \mathrm{NO}_2 \underset{K_2}{\stackrel{K_1}{\rightleftharpoons}} N_2 \mathrm{O}_4$
the rate of disappearance of $\mathrm{NO}_{2}$ is equal to

276094 For the following cell reaction,
$\mathrm{Ag}\left \vert\mathrm{Ag}^{+}\right \vert \mathrm{AgCl}\left \vert\mathrm{Cl}^{-}\right \vert \mathrm{Cl}_{2}, \mathrm{Pt}$
$\Delta G_{\mathrm{f}}^{\circ}(\mathrm{AgCl})=-109 \mathrm{~kJ} / \mathrm{mol}$
$\Delta G_{\mathrm{f}}^{\circ}\left(\mathrm{Cl}^{-}\right)=-129 \mathrm{~kJ} / \mathrm{mol}$
$\Delta G_{\mathrm{f}}^{\circ}\left(\mathrm{Ag}^{+}\right)=78 \mathrm{~kJ} / \mathrm{mol}$
$\mathrm{E}^{\circ}$ of the cell is

276095 The standard free energy change of a reaction is $\Delta G^{\circ}=-115 \mathrm{~kJ}$ at $298 \mathrm{~K}$. Calculate the equilibrium constant $k_{p}$ in $\log k_{p}$ $\left(\mathrm{R}=\mathbf{8 . 3 1 4} \mathrm{Jk}^{-1} \mathrm{~mol}^{-1}\right)$.

276091 The resistance of $0.01 \mathrm{~N}$ solution of an electrolyte was found to be $220 \mathrm{ohm}$ at $298 \mathrm{~K}$ using a conductivity cell with a cell constant of $0.88 \mathrm{~cm}^{-1}$. The value of equivalent conductance of solution is -

276092 The EMF of the cell,
$\mathrm{Mg}\left \vert\mathrm{Mg}^{2+}(0.01 \mathrm{M})\right \vert\left \vert\mathrm{Sn}^{2+}(0.1 \mathrm{M})\right \vert \mathrm{Sn}$ at $298 \mathrm{~K}$ is
$\left(\mathrm{E}_{\mathrm{Mg}^{2+} / \mathrm{Mg}}^{\circ}=-2.34 \mathrm{~V}, \mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{\circ}=-0.14 \mathrm{~V}\right)$

276093 In the reversible reaction.
$2 \mathrm{NO}_2 \underset{K_2}{\stackrel{K_1}{\rightleftharpoons}} N_2 \mathrm{O}_4$
the rate of disappearance of $\mathrm{NO}_{2}$ is equal to

276094 For the following cell reaction,
$\mathrm{Ag}\left \vert\mathrm{Ag}^{+}\right \vert \mathrm{AgCl}\left \vert\mathrm{Cl}^{-}\right \vert \mathrm{Cl}_{2}, \mathrm{Pt}$
$\Delta G_{\mathrm{f}}^{\circ}(\mathrm{AgCl})=-109 \mathrm{~kJ} / \mathrm{mol}$
$\Delta G_{\mathrm{f}}^{\circ}\left(\mathrm{Cl}^{-}\right)=-129 \mathrm{~kJ} / \mathrm{mol}$
$\Delta G_{\mathrm{f}}^{\circ}\left(\mathrm{Ag}^{+}\right)=78 \mathrm{~kJ} / \mathrm{mol}$
$\mathrm{E}^{\circ}$ of the cell is

276095 The standard free energy change of a reaction is $\Delta G^{\circ}=-115 \mathrm{~kJ}$ at $298 \mathrm{~K}$. Calculate the equilibrium constant $k_{p}$ in $\log k_{p}$ $\left(\mathrm{R}=\mathbf{8 . 3 1 4} \mathrm{Jk}^{-1} \mathrm{~mol}^{-1}\right)$.

276091 The resistance of $0.01 \mathrm{~N}$ solution of an electrolyte was found to be $220 \mathrm{ohm}$ at $298 \mathrm{~K}$ using a conductivity cell with a cell constant of $0.88 \mathrm{~cm}^{-1}$. The value of equivalent conductance of solution is -

276092 The EMF of the cell,
$\mathrm{Mg}\left \vert\mathrm{Mg}^{2+}(0.01 \mathrm{M})\right \vert\left \vert\mathrm{Sn}^{2+}(0.1 \mathrm{M})\right \vert \mathrm{Sn}$ at $298 \mathrm{~K}$ is
$\left(\mathrm{E}_{\mathrm{Mg}^{2+} / \mathrm{Mg}}^{\circ}=-2.34 \mathrm{~V}, \mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{\circ}=-0.14 \mathrm{~V}\right)$

276093 In the reversible reaction.
$2 \mathrm{NO}_2 \underset{K_2}{\stackrel{K_1}{\rightleftharpoons}} N_2 \mathrm{O}_4$
the rate of disappearance of $\mathrm{NO}_{2}$ is equal to

276094 For the following cell reaction,
$\mathrm{Ag}\left \vert\mathrm{Ag}^{+}\right \vert \mathrm{AgCl}\left \vert\mathrm{Cl}^{-}\right \vert \mathrm{Cl}_{2}, \mathrm{Pt}$
$\Delta G_{\mathrm{f}}^{\circ}(\mathrm{AgCl})=-109 \mathrm{~kJ} / \mathrm{mol}$
$\Delta G_{\mathrm{f}}^{\circ}\left(\mathrm{Cl}^{-}\right)=-129 \mathrm{~kJ} / \mathrm{mol}$
$\Delta G_{\mathrm{f}}^{\circ}\left(\mathrm{Ag}^{+}\right)=78 \mathrm{~kJ} / \mathrm{mol}$
$\mathrm{E}^{\circ}$ of the cell is

276095 The standard free energy change of a reaction is $\Delta G^{\circ}=-115 \mathrm{~kJ}$ at $298 \mathrm{~K}$. Calculate the equilibrium constant $k_{p}$ in $\log k_{p}$ $\left(\mathrm{R}=\mathbf{8 . 3 1 4} \mathrm{Jk}^{-1} \mathrm{~mol}^{-1}\right)$.

276091 The resistance of $0.01 \mathrm{~N}$ solution of an electrolyte was found to be $220 \mathrm{ohm}$ at $298 \mathrm{~K}$ using a conductivity cell with a cell constant of $0.88 \mathrm{~cm}^{-1}$. The value of equivalent conductance of solution is -

276092 The EMF of the cell,
$\mathrm{Mg}\left \vert\mathrm{Mg}^{2+}(0.01 \mathrm{M})\right \vert\left \vert\mathrm{Sn}^{2+}(0.1 \mathrm{M})\right \vert \mathrm{Sn}$ at $298 \mathrm{~K}$ is
$\left(\mathrm{E}_{\mathrm{Mg}^{2+} / \mathrm{Mg}}^{\circ}=-2.34 \mathrm{~V}, \mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{\circ}=-0.14 \mathrm{~V}\right)$

276093 In the reversible reaction.
$2 \mathrm{NO}_2 \underset{K_2}{\stackrel{K_1}{\rightleftharpoons}} N_2 \mathrm{O}_4$
the rate of disappearance of $\mathrm{NO}_{2}$ is equal to

276094 For the following cell reaction,
$\mathrm{Ag}\left \vert\mathrm{Ag}^{+}\right \vert \mathrm{AgCl}\left \vert\mathrm{Cl}^{-}\right \vert \mathrm{Cl}_{2}, \mathrm{Pt}$
$\Delta G_{\mathrm{f}}^{\circ}(\mathrm{AgCl})=-109 \mathrm{~kJ} / \mathrm{mol}$
$\Delta G_{\mathrm{f}}^{\circ}\left(\mathrm{Cl}^{-}\right)=-129 \mathrm{~kJ} / \mathrm{mol}$
$\Delta G_{\mathrm{f}}^{\circ}\left(\mathrm{Ag}^{+}\right)=78 \mathrm{~kJ} / \mathrm{mol}$
$\mathrm{E}^{\circ}$ of the cell is

276095 The standard free energy change of a reaction is $\Delta G^{\circ}=-115 \mathrm{~kJ}$ at $298 \mathrm{~K}$. Calculate the equilibrium constant $k_{p}$ in $\log k_{p}$ $\left(\mathrm{R}=\mathbf{8 . 3 1 4} \mathrm{Jk}^{-1} \mathrm{~mol}^{-1}\right)$.

276091 The resistance of $0.01 \mathrm{~N}$ solution of an electrolyte was found to be $220 \mathrm{ohm}$ at $298 \mathrm{~K}$ using a conductivity cell with a cell constant of $0.88 \mathrm{~cm}^{-1}$. The value of equivalent conductance of solution is -

276092 The EMF of the cell,
$\mathrm{Mg}\left \vert\mathrm{Mg}^{2+}(0.01 \mathrm{M})\right \vert\left \vert\mathrm{Sn}^{2+}(0.1 \mathrm{M})\right \vert \mathrm{Sn}$ at $298 \mathrm{~K}$ is
$\left(\mathrm{E}_{\mathrm{Mg}^{2+} / \mathrm{Mg}}^{\circ}=-2.34 \mathrm{~V}, \mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{\circ}=-0.14 \mathrm{~V}\right)$

276093 In the reversible reaction.
$2 \mathrm{NO}_2 \underset{K_2}{\stackrel{K_1}{\rightleftharpoons}} N_2 \mathrm{O}_4$
the rate of disappearance of $\mathrm{NO}_{2}$ is equal to

276094 For the following cell reaction,
$\mathrm{Ag}\left \vert\mathrm{Ag}^{+}\right \vert \mathrm{AgCl}\left \vert\mathrm{Cl}^{-}\right \vert \mathrm{Cl}_{2}, \mathrm{Pt}$
$\Delta G_{\mathrm{f}}^{\circ}(\mathrm{AgCl})=-109 \mathrm{~kJ} / \mathrm{mol}$
$\Delta G_{\mathrm{f}}^{\circ}\left(\mathrm{Cl}^{-}\right)=-129 \mathrm{~kJ} / \mathrm{mol}$
$\Delta G_{\mathrm{f}}^{\circ}\left(\mathrm{Ag}^{+}\right)=78 \mathrm{~kJ} / \mathrm{mol}$
$\mathrm{E}^{\circ}$ of the cell is

276095 The standard free energy change of a reaction is $\Delta G^{\circ}=-115 \mathrm{~kJ}$ at $298 \mathrm{~K}$. Calculate the equilibrium constant $k_{p}$ in $\log k_{p}$ $\left(\mathrm{R}=\mathbf{8 . 3 1 4} \mathrm{Jk}^{-1} \mathrm{~mol}^{-1}\right)$.