According to Le-Chatelier's principal, when $\Delta \mathrm{n}_{\mathrm{g}}=0$ there is no effect of pressure and volume change on the equilibrium So, (a) $\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}), \Delta \mathrm{n}_{\mathrm{g}}=0$ (b) $\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}), \Delta \mathrm{n}_{\mathrm{g}}=-2$ (c) $\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}), \Delta \mathrm{n}_{\mathrm{g}}=1$ (d) $2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}), \Delta \mathrm{n}_{\mathrm{g}}=-1$ Hence, $\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$ is correct answer.
J and K CET-(2006)
Chemical Equilibrium
229193
For $\mathrm{NaC1}$, the $K_{\mathrm{sp}}=36 \mathrm{~mol}^{2} \mathrm{~L}^{-2}$, the molar concentration of it will be
229196
If pressure increases then its effect on given equilibrium $2NO \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})$ is shift in:
1 Forward direction
2 Backward direction
3 No effect
4 None of the above
Explanation:
The reaction is- $\begin{aligned} & 2 \mathrm{NO}(\mathrm{g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \\ & \Delta \mathrm{n}_{\mathrm{g}}=2-2=0 \end{aligned}$ $\because$ There is no change in number of moles of reactant and products. $\therefore$ The reaction is not effected by change in pressure.
BCECE-2005
Chemical Equilibrium
229201
Of the following which change will shift the reaction towards the product? $\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons \quad 2 \mathrm{I}(\mathrm{g}), \Delta \mathrm{H}_{\mathrm{r}}^{0}(298 \mathrm{~K})=+150 \mathrm{~kJ}$
1 Increase in concentration of $\mathrm{I}_{2}$
2 Decrease in concentration of $\mathrm{I}_{2}$
3 Increase in temperature
4 Increase in total pressure
Explanation:
The reaction is $\begin{aligned} & \mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g}) \\ & \Delta \mathrm{H}_{\mathrm{r}}^{0}(298 \mathrm{~K})=+150 \mathrm{~kJ} . \end{aligned}$ According to Le-chateliers principal, the condition favouring the formation of I are given below. (i) High concentration of $\mathrm{I}_{2}(\mathrm{~g})$ (ii) High temperature (iii) Low pressure Hence option (c) is correct.
According to Le-Chatelier's principal, when $\Delta \mathrm{n}_{\mathrm{g}}=0$ there is no effect of pressure and volume change on the equilibrium So, (a) $\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}), \Delta \mathrm{n}_{\mathrm{g}}=0$ (b) $\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}), \Delta \mathrm{n}_{\mathrm{g}}=-2$ (c) $\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}), \Delta \mathrm{n}_{\mathrm{g}}=1$ (d) $2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}), \Delta \mathrm{n}_{\mathrm{g}}=-1$ Hence, $\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$ is correct answer.
J and K CET-(2006)
Chemical Equilibrium
229193
For $\mathrm{NaC1}$, the $K_{\mathrm{sp}}=36 \mathrm{~mol}^{2} \mathrm{~L}^{-2}$, the molar concentration of it will be
229196
If pressure increases then its effect on given equilibrium $2NO \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})$ is shift in:
1 Forward direction
2 Backward direction
3 No effect
4 None of the above
Explanation:
The reaction is- $\begin{aligned} & 2 \mathrm{NO}(\mathrm{g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \\ & \Delta \mathrm{n}_{\mathrm{g}}=2-2=0 \end{aligned}$ $\because$ There is no change in number of moles of reactant and products. $\therefore$ The reaction is not effected by change in pressure.
BCECE-2005
Chemical Equilibrium
229201
Of the following which change will shift the reaction towards the product? $\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons \quad 2 \mathrm{I}(\mathrm{g}), \Delta \mathrm{H}_{\mathrm{r}}^{0}(298 \mathrm{~K})=+150 \mathrm{~kJ}$
1 Increase in concentration of $\mathrm{I}_{2}$
2 Decrease in concentration of $\mathrm{I}_{2}$
3 Increase in temperature
4 Increase in total pressure
Explanation:
The reaction is $\begin{aligned} & \mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g}) \\ & \Delta \mathrm{H}_{\mathrm{r}}^{0}(298 \mathrm{~K})=+150 \mathrm{~kJ} . \end{aligned}$ According to Le-chateliers principal, the condition favouring the formation of I are given below. (i) High concentration of $\mathrm{I}_{2}(\mathrm{~g})$ (ii) High temperature (iii) Low pressure Hence option (c) is correct.
According to Le-Chatelier's principal, when $\Delta \mathrm{n}_{\mathrm{g}}=0$ there is no effect of pressure and volume change on the equilibrium So, (a) $\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}), \Delta \mathrm{n}_{\mathrm{g}}=0$ (b) $\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}), \Delta \mathrm{n}_{\mathrm{g}}=-2$ (c) $\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}), \Delta \mathrm{n}_{\mathrm{g}}=1$ (d) $2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}), \Delta \mathrm{n}_{\mathrm{g}}=-1$ Hence, $\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$ is correct answer.
J and K CET-(2006)
Chemical Equilibrium
229193
For $\mathrm{NaC1}$, the $K_{\mathrm{sp}}=36 \mathrm{~mol}^{2} \mathrm{~L}^{-2}$, the molar concentration of it will be
229196
If pressure increases then its effect on given equilibrium $2NO \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})$ is shift in:
1 Forward direction
2 Backward direction
3 No effect
4 None of the above
Explanation:
The reaction is- $\begin{aligned} & 2 \mathrm{NO}(\mathrm{g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \\ & \Delta \mathrm{n}_{\mathrm{g}}=2-2=0 \end{aligned}$ $\because$ There is no change in number of moles of reactant and products. $\therefore$ The reaction is not effected by change in pressure.
BCECE-2005
Chemical Equilibrium
229201
Of the following which change will shift the reaction towards the product? $\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons \quad 2 \mathrm{I}(\mathrm{g}), \Delta \mathrm{H}_{\mathrm{r}}^{0}(298 \mathrm{~K})=+150 \mathrm{~kJ}$
1 Increase in concentration of $\mathrm{I}_{2}$
2 Decrease in concentration of $\mathrm{I}_{2}$
3 Increase in temperature
4 Increase in total pressure
Explanation:
The reaction is $\begin{aligned} & \mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g}) \\ & \Delta \mathrm{H}_{\mathrm{r}}^{0}(298 \mathrm{~K})=+150 \mathrm{~kJ} . \end{aligned}$ According to Le-chateliers principal, the condition favouring the formation of I are given below. (i) High concentration of $\mathrm{I}_{2}(\mathrm{~g})$ (ii) High temperature (iii) Low pressure Hence option (c) is correct.
According to Le-Chatelier's principal, when $\Delta \mathrm{n}_{\mathrm{g}}=0$ there is no effect of pressure and volume change on the equilibrium So, (a) $\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}), \Delta \mathrm{n}_{\mathrm{g}}=0$ (b) $\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}), \Delta \mathrm{n}_{\mathrm{g}}=-2$ (c) $\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}), \Delta \mathrm{n}_{\mathrm{g}}=1$ (d) $2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}), \Delta \mathrm{n}_{\mathrm{g}}=-1$ Hence, $\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$ is correct answer.
J and K CET-(2006)
Chemical Equilibrium
229193
For $\mathrm{NaC1}$, the $K_{\mathrm{sp}}=36 \mathrm{~mol}^{2} \mathrm{~L}^{-2}$, the molar concentration of it will be
229196
If pressure increases then its effect on given equilibrium $2NO \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})$ is shift in:
1 Forward direction
2 Backward direction
3 No effect
4 None of the above
Explanation:
The reaction is- $\begin{aligned} & 2 \mathrm{NO}(\mathrm{g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \\ & \Delta \mathrm{n}_{\mathrm{g}}=2-2=0 \end{aligned}$ $\because$ There is no change in number of moles of reactant and products. $\therefore$ The reaction is not effected by change in pressure.
BCECE-2005
Chemical Equilibrium
229201
Of the following which change will shift the reaction towards the product? $\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons \quad 2 \mathrm{I}(\mathrm{g}), \Delta \mathrm{H}_{\mathrm{r}}^{0}(298 \mathrm{~K})=+150 \mathrm{~kJ}$
1 Increase in concentration of $\mathrm{I}_{2}$
2 Decrease in concentration of $\mathrm{I}_{2}$
3 Increase in temperature
4 Increase in total pressure
Explanation:
The reaction is $\begin{aligned} & \mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g}) \\ & \Delta \mathrm{H}_{\mathrm{r}}^{0}(298 \mathrm{~K})=+150 \mathrm{~kJ} . \end{aligned}$ According to Le-chateliers principal, the condition favouring the formation of I are given below. (i) High concentration of $\mathrm{I}_{2}(\mathrm{~g})$ (ii) High temperature (iii) Low pressure Hence option (c) is correct.
