An increase in pressure favours the reaction in the direction that produced smaller number of moles of gases. As in reaction (b), 3 moles of reactants produces lesser i.e. 2 mole of products. Therefore, the reaction will move in forward direction.
J and K CET-(2013)
Chemical Equilibrium
229174
In which reaction will an increase in the volume of the container favour the formation of products?
As we increase the volume of the container the concentration decreases of all the gaseous states in the reaction by Le-Chatelier's principle, the reaction will move in the direction of increasing no. of moles.
SCRA 2012
Chemical Equilibrium
229175
According to Le-Chartelier's principle, the equilibrium constant of a reversible reaction will not shift by
1 Increasing the temperature of an exothermic reaction
2 Increasing the temperature of an endothermic reaction
3 Charging the concentrations of the reactants
4 The effect of a catalyst.
Explanation:
According to Le-Chatelier's principal, the equilibrium state of a reversible reaction will not shift by the effect of a catalyst but equilibrium constant remain unaffected by the effect of catalyst as well as change in concentration of the reactants.
J and K CET-(2012)
Chemical Equilibrium
229178
$\mathrm{H}_{2} \mathrm{~S}$ gas when passed through a solution of cations containing $\mathrm{HCl}$ precipitates the cations of second group of qualitative analysis but not those belonging to the fourth group. It is because
1 Presence of $\mathrm{HCl}$ decreases the sulphide ion concentration.
2 Solubility product of group II sulphides is more than that of group IV sulphides.
3 Presence of $\mathrm{HCl}$ increases the sulphide ion concentration.
4 Sulphides of group IV cations are unstable in $\mathrm{HCl}$.
Explanation:
$\mathrm{H}_{2} \mathrm{~S}$ gas is passed in presence of $\mathrm{HCl}$, in qualitative analysis of cation of second group, therefore, due to common ion effect, lower concentration of sulphide ions is obtained which is sufficient for the precipitation of second group cations in form of their sulphides due to lower value of their solubility product $\left(\mathrm{K}_{\mathrm{sp}}\right)$. Here, fourth group cations are not precipitated because they require for exceeding their ionic product to their solubility products and higher sulphide ions concentration due to their higher $K_{s p}$ which is not obtained here due to common ion effect.
An increase in pressure favours the reaction in the direction that produced smaller number of moles of gases. As in reaction (b), 3 moles of reactants produces lesser i.e. 2 mole of products. Therefore, the reaction will move in forward direction.
J and K CET-(2013)
Chemical Equilibrium
229174
In which reaction will an increase in the volume of the container favour the formation of products?
As we increase the volume of the container the concentration decreases of all the gaseous states in the reaction by Le-Chatelier's principle, the reaction will move in the direction of increasing no. of moles.
SCRA 2012
Chemical Equilibrium
229175
According to Le-Chartelier's principle, the equilibrium constant of a reversible reaction will not shift by
1 Increasing the temperature of an exothermic reaction
2 Increasing the temperature of an endothermic reaction
3 Charging the concentrations of the reactants
4 The effect of a catalyst.
Explanation:
According to Le-Chatelier's principal, the equilibrium state of a reversible reaction will not shift by the effect of a catalyst but equilibrium constant remain unaffected by the effect of catalyst as well as change in concentration of the reactants.
J and K CET-(2012)
Chemical Equilibrium
229178
$\mathrm{H}_{2} \mathrm{~S}$ gas when passed through a solution of cations containing $\mathrm{HCl}$ precipitates the cations of second group of qualitative analysis but not those belonging to the fourth group. It is because
1 Presence of $\mathrm{HCl}$ decreases the sulphide ion concentration.
2 Solubility product of group II sulphides is more than that of group IV sulphides.
3 Presence of $\mathrm{HCl}$ increases the sulphide ion concentration.
4 Sulphides of group IV cations are unstable in $\mathrm{HCl}$.
Explanation:
$\mathrm{H}_{2} \mathrm{~S}$ gas is passed in presence of $\mathrm{HCl}$, in qualitative analysis of cation of second group, therefore, due to common ion effect, lower concentration of sulphide ions is obtained which is sufficient for the precipitation of second group cations in form of their sulphides due to lower value of their solubility product $\left(\mathrm{K}_{\mathrm{sp}}\right)$. Here, fourth group cations are not precipitated because they require for exceeding their ionic product to their solubility products and higher sulphide ions concentration due to their higher $K_{s p}$ which is not obtained here due to common ion effect.
An increase in pressure favours the reaction in the direction that produced smaller number of moles of gases. As in reaction (b), 3 moles of reactants produces lesser i.e. 2 mole of products. Therefore, the reaction will move in forward direction.
J and K CET-(2013)
Chemical Equilibrium
229174
In which reaction will an increase in the volume of the container favour the formation of products?
As we increase the volume of the container the concentration decreases of all the gaseous states in the reaction by Le-Chatelier's principle, the reaction will move in the direction of increasing no. of moles.
SCRA 2012
Chemical Equilibrium
229175
According to Le-Chartelier's principle, the equilibrium constant of a reversible reaction will not shift by
1 Increasing the temperature of an exothermic reaction
2 Increasing the temperature of an endothermic reaction
3 Charging the concentrations of the reactants
4 The effect of a catalyst.
Explanation:
According to Le-Chatelier's principal, the equilibrium state of a reversible reaction will not shift by the effect of a catalyst but equilibrium constant remain unaffected by the effect of catalyst as well as change in concentration of the reactants.
J and K CET-(2012)
Chemical Equilibrium
229178
$\mathrm{H}_{2} \mathrm{~S}$ gas when passed through a solution of cations containing $\mathrm{HCl}$ precipitates the cations of second group of qualitative analysis but not those belonging to the fourth group. It is because
1 Presence of $\mathrm{HCl}$ decreases the sulphide ion concentration.
2 Solubility product of group II sulphides is more than that of group IV sulphides.
3 Presence of $\mathrm{HCl}$ increases the sulphide ion concentration.
4 Sulphides of group IV cations are unstable in $\mathrm{HCl}$.
Explanation:
$\mathrm{H}_{2} \mathrm{~S}$ gas is passed in presence of $\mathrm{HCl}$, in qualitative analysis of cation of second group, therefore, due to common ion effect, lower concentration of sulphide ions is obtained which is sufficient for the precipitation of second group cations in form of their sulphides due to lower value of their solubility product $\left(\mathrm{K}_{\mathrm{sp}}\right)$. Here, fourth group cations are not precipitated because they require for exceeding their ionic product to their solubility products and higher sulphide ions concentration due to their higher $K_{s p}$ which is not obtained here due to common ion effect.
An increase in pressure favours the reaction in the direction that produced smaller number of moles of gases. As in reaction (b), 3 moles of reactants produces lesser i.e. 2 mole of products. Therefore, the reaction will move in forward direction.
J and K CET-(2013)
Chemical Equilibrium
229174
In which reaction will an increase in the volume of the container favour the formation of products?
As we increase the volume of the container the concentration decreases of all the gaseous states in the reaction by Le-Chatelier's principle, the reaction will move in the direction of increasing no. of moles.
SCRA 2012
Chemical Equilibrium
229175
According to Le-Chartelier's principle, the equilibrium constant of a reversible reaction will not shift by
1 Increasing the temperature of an exothermic reaction
2 Increasing the temperature of an endothermic reaction
3 Charging the concentrations of the reactants
4 The effect of a catalyst.
Explanation:
According to Le-Chatelier's principal, the equilibrium state of a reversible reaction will not shift by the effect of a catalyst but equilibrium constant remain unaffected by the effect of catalyst as well as change in concentration of the reactants.
J and K CET-(2012)
Chemical Equilibrium
229178
$\mathrm{H}_{2} \mathrm{~S}$ gas when passed through a solution of cations containing $\mathrm{HCl}$ precipitates the cations of second group of qualitative analysis but not those belonging to the fourth group. It is because
1 Presence of $\mathrm{HCl}$ decreases the sulphide ion concentration.
2 Solubility product of group II sulphides is more than that of group IV sulphides.
3 Presence of $\mathrm{HCl}$ increases the sulphide ion concentration.
4 Sulphides of group IV cations are unstable in $\mathrm{HCl}$.
Explanation:
$\mathrm{H}_{2} \mathrm{~S}$ gas is passed in presence of $\mathrm{HCl}$, in qualitative analysis of cation of second group, therefore, due to common ion effect, lower concentration of sulphide ions is obtained which is sufficient for the precipitation of second group cations in form of their sulphides due to lower value of their solubility product $\left(\mathrm{K}_{\mathrm{sp}}\right)$. Here, fourth group cations are not precipitated because they require for exceeding their ionic product to their solubility products and higher sulphide ions concentration due to their higher $K_{s p}$ which is not obtained here due to common ion effect.
