Classify equilibrium on the basis of physical state of its components. 

Graph of a reversible process;

${\mathrm{N}}_2+3{\mathrm{H}}_2 \stackrel{}{\rightleftharpoons } 2{\mathrm{NH}}_3+\mathrm{Heat}$

is given. Analyse the graph and answer the following question.

Identify the part of the graph which represents the forward reaction

[ OA, BA, AC]

The value of ${\mathrm{K}}_{\mathrm{c}} \mathrm{us} 64 \mathrm{at} 800 \mathrm{K}$ for the reaction

${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\to 2 {\mathrm{NH}}_3 \left(\mathrm{g}\right)$

The value of ${\mathrm{K}}_{\mathrm{c}}$ for the following reaction is :

${\mathrm{NH}}_3 \left(\mathrm{g}\right)\to \frac{1}{2} {\mathrm{N}}_2 \left(\mathrm{g}\right)+\frac{3}{2} {\mathrm{H}}_2 \left(\mathrm{g}\right)$

The reaction

${\mathrm{CaCO}}_3\left(\mathrm{s}\right)\rightleftharpoons \mathrm{CaO}\left(\mathrm{s}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)$

is in equilibrium in a closed vessel at $298 \mathrm{K}.$ The partial pressure (in atm) of ${\mathrm{CO}}_2\left(\mathrm{g}\right)$ in the reaction vessel is closest to:

[Given: The change in Gibbs energies of formation at $298 \mathrm{K}$ and $1$ bar for

$\mathrm{CaO}\left(\mathrm{s}\right)=-603.501 \mathrm{kJ} {\mathrm{mol}}^{-1}$

${\mathrm{CO}}_2\left(\mathrm{g}\right)=-394.389 \mathrm{kJ} {\mathrm{mol}}^{-1}$

${\mathrm{CaCO}}_3\left(\mathrm{s}\right)=-1128.79 \mathrm{kJ} {\mathrm{mol}}^{-1}$

Gas constant $\mathrm{R}=8.314 \mathrm{J} {\mathrm{K}}^{-1} {\mathrm{mol}}^{-1}$]

Using the data provided, find the value of equilibrium constant for the following reaction at $298 \mathrm{K}$ and $1$ atm pressure. ${\mathrm{NO}}_{\left(\mathrm{g}\right)}+\frac{1}{2}{\mathrm{O}}_2( \mathrm{g})\rightleftharpoons {\mathrm{NO}}_2( \mathrm{g})$
${\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{H}}^0\left({\mathrm{NO}}_{\left(\mathrm{g}\right)}\right)=90.4\mathrm{kJ}.{\mathrm{mol}}^{-1}$

${\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{H}}^0\left({\mathrm{NO}}_2\left(\mathrm{g}\right)\right)=32.48\mathrm{kJ}\cdot {\mathrm{mol}}^{-1}$

${\mathrm{\Delta S}}^{\circ }\mathrm{at} 298\mathrm{K}=-70.8\mathrm{J}\cdot {\mathrm{K}}^{-1}\cdot {\mathrm{mol}}^{-1}$

$\mathrm{antilog}(0.50)=3162$

Graph of a reversible process,

is given. Analyse the graph and answer the following question.

From the given statements, select the correct ones regarding chemical equilibrium.

(i0 The chemical equilibrium is 'static' at the molecular level.

(ii) Both reactants and products co-exist.

(iii) The rates of forward reaction and backward reactions are equal.

(iv) Chemical equilibrium is attained in an open system.

K_p for the reaction
${\mathrm{CuSO}}_4·3{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{s}\right)}+2{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}\to {\mathrm{CuSO}}_4·5{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{s}\right)}$   is $1·0\times 10^4 {\text{atm}}^{-2}$  at certain  temperature. What lowest relative humidity of air can be achieved using $CuS{O}_4.3H_{2}O$ as drying agent at that temperature ?
Aqueous tension at the given temperature = 16 torr