Order of Reaction-No. of reactant molecules taking part in the reaction/no. of reactant molecules that effects the rate of reaction. $\mathrm{NH}_4 \mathrm{NO}_2 \rightarrow \mathrm{N}_2+2 \mathrm{H}_2 \mathrm{O}$ Rate of equation $\mathrm{K}\left[\mathrm{NH}_4 \mathrm{NO}_2\right]^1$ $\therefore$ Order of this reaction is (1) Hence, option (a) is correct.
CG PET-2010
Thermodynamics
272691
The incorrect expression among the following is
4 For isothermal process, $\mathrm{W}_{\text {reversible }}=-n R T \ln \frac{V_f}{\mathrm{~V}_{\mathrm{i}}}$
Explanation:
We know that, $\Delta \mathrm{G}^{\circ}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}$ Where $\Delta \mathrm{G}^{\circ}$ is standard $\mathrm{Gibbs}$ free energy change $\Delta \mathrm{H}^{\circ}$ is standard Enthalpy change, $\mathrm{T} \Delta \mathrm{S}^{\circ}$ is standard Entropy change, $\mathrm{T}$ is temperature. $\Delta \mathrm{G}^0=-\mathrm{RT} \times \ln (\mathrm{K})$, where $\mathrm{K}$ is the equilibrium constant $-\mathrm{RT} \times \ln (\mathrm{K})=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}$ $\operatorname{In}(\mathrm{K})=-\left[\frac{\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}}{\mathrm{RT}}\right]$
272693
The internal energy change when a system goes from state $A$ to $B$ is $40 \mathrm{~kJ} / \mathrm{mol}$. If the system goes from $A$ to $B$ by a reversible path and returns to state $A$ by an irreversible path, what would be the net change in internal energy?
1 $40 \mathrm{~kJ}$
2 $>40 \mathrm{~kJ}$
3 $<40 \mathrm{~kJ}$
4 zero
Explanation:
For a cyclic process the net change in the internal energy is zero because the change in internal energy does not depend on the path. $\because \quad \Delta \mathrm{U}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}$ When system returns to its initial state, $\therefore \quad \Delta \mathrm{U}=0$
Order of Reaction-No. of reactant molecules taking part in the reaction/no. of reactant molecules that effects the rate of reaction. $\mathrm{NH}_4 \mathrm{NO}_2 \rightarrow \mathrm{N}_2+2 \mathrm{H}_2 \mathrm{O}$ Rate of equation $\mathrm{K}\left[\mathrm{NH}_4 \mathrm{NO}_2\right]^1$ $\therefore$ Order of this reaction is (1) Hence, option (a) is correct.
CG PET-2010
Thermodynamics
272691
The incorrect expression among the following is
4 For isothermal process, $\mathrm{W}_{\text {reversible }}=-n R T \ln \frac{V_f}{\mathrm{~V}_{\mathrm{i}}}$
Explanation:
We know that, $\Delta \mathrm{G}^{\circ}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}$ Where $\Delta \mathrm{G}^{\circ}$ is standard $\mathrm{Gibbs}$ free energy change $\Delta \mathrm{H}^{\circ}$ is standard Enthalpy change, $\mathrm{T} \Delta \mathrm{S}^{\circ}$ is standard Entropy change, $\mathrm{T}$ is temperature. $\Delta \mathrm{G}^0=-\mathrm{RT} \times \ln (\mathrm{K})$, where $\mathrm{K}$ is the equilibrium constant $-\mathrm{RT} \times \ln (\mathrm{K})=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}$ $\operatorname{In}(\mathrm{K})=-\left[\frac{\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}}{\mathrm{RT}}\right]$
272693
The internal energy change when a system goes from state $A$ to $B$ is $40 \mathrm{~kJ} / \mathrm{mol}$. If the system goes from $A$ to $B$ by a reversible path and returns to state $A$ by an irreversible path, what would be the net change in internal energy?
1 $40 \mathrm{~kJ}$
2 $>40 \mathrm{~kJ}$
3 $<40 \mathrm{~kJ}$
4 zero
Explanation:
For a cyclic process the net change in the internal energy is zero because the change in internal energy does not depend on the path. $\because \quad \Delta \mathrm{U}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}$ When system returns to its initial state, $\therefore \quad \Delta \mathrm{U}=0$
Order of Reaction-No. of reactant molecules taking part in the reaction/no. of reactant molecules that effects the rate of reaction. $\mathrm{NH}_4 \mathrm{NO}_2 \rightarrow \mathrm{N}_2+2 \mathrm{H}_2 \mathrm{O}$ Rate of equation $\mathrm{K}\left[\mathrm{NH}_4 \mathrm{NO}_2\right]^1$ $\therefore$ Order of this reaction is (1) Hence, option (a) is correct.
CG PET-2010
Thermodynamics
272691
The incorrect expression among the following is
4 For isothermal process, $\mathrm{W}_{\text {reversible }}=-n R T \ln \frac{V_f}{\mathrm{~V}_{\mathrm{i}}}$
Explanation:
We know that, $\Delta \mathrm{G}^{\circ}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}$ Where $\Delta \mathrm{G}^{\circ}$ is standard $\mathrm{Gibbs}$ free energy change $\Delta \mathrm{H}^{\circ}$ is standard Enthalpy change, $\mathrm{T} \Delta \mathrm{S}^{\circ}$ is standard Entropy change, $\mathrm{T}$ is temperature. $\Delta \mathrm{G}^0=-\mathrm{RT} \times \ln (\mathrm{K})$, where $\mathrm{K}$ is the equilibrium constant $-\mathrm{RT} \times \ln (\mathrm{K})=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}$ $\operatorname{In}(\mathrm{K})=-\left[\frac{\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}}{\mathrm{RT}}\right]$
272693
The internal energy change when a system goes from state $A$ to $B$ is $40 \mathrm{~kJ} / \mathrm{mol}$. If the system goes from $A$ to $B$ by a reversible path and returns to state $A$ by an irreversible path, what would be the net change in internal energy?
1 $40 \mathrm{~kJ}$
2 $>40 \mathrm{~kJ}$
3 $<40 \mathrm{~kJ}$
4 zero
Explanation:
For a cyclic process the net change in the internal energy is zero because the change in internal energy does not depend on the path. $\because \quad \Delta \mathrm{U}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}$ When system returns to its initial state, $\therefore \quad \Delta \mathrm{U}=0$
Order of Reaction-No. of reactant molecules taking part in the reaction/no. of reactant molecules that effects the rate of reaction. $\mathrm{NH}_4 \mathrm{NO}_2 \rightarrow \mathrm{N}_2+2 \mathrm{H}_2 \mathrm{O}$ Rate of equation $\mathrm{K}\left[\mathrm{NH}_4 \mathrm{NO}_2\right]^1$ $\therefore$ Order of this reaction is (1) Hence, option (a) is correct.
CG PET-2010
Thermodynamics
272691
The incorrect expression among the following is
4 For isothermal process, $\mathrm{W}_{\text {reversible }}=-n R T \ln \frac{V_f}{\mathrm{~V}_{\mathrm{i}}}$
Explanation:
We know that, $\Delta \mathrm{G}^{\circ}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}$ Where $\Delta \mathrm{G}^{\circ}$ is standard $\mathrm{Gibbs}$ free energy change $\Delta \mathrm{H}^{\circ}$ is standard Enthalpy change, $\mathrm{T} \Delta \mathrm{S}^{\circ}$ is standard Entropy change, $\mathrm{T}$ is temperature. $\Delta \mathrm{G}^0=-\mathrm{RT} \times \ln (\mathrm{K})$, where $\mathrm{K}$ is the equilibrium constant $-\mathrm{RT} \times \ln (\mathrm{K})=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}$ $\operatorname{In}(\mathrm{K})=-\left[\frac{\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ}}{\mathrm{RT}}\right]$
272693
The internal energy change when a system goes from state $A$ to $B$ is $40 \mathrm{~kJ} / \mathrm{mol}$. If the system goes from $A$ to $B$ by a reversible path and returns to state $A$ by an irreversible path, what would be the net change in internal energy?
1 $40 \mathrm{~kJ}$
2 $>40 \mathrm{~kJ}$
3 $<40 \mathrm{~kJ}$
4 zero
Explanation:
For a cyclic process the net change in the internal energy is zero because the change in internal energy does not depend on the path. $\because \quad \Delta \mathrm{U}=\mathrm{U}_{\mathrm{f}}-\mathrm{U}_{\mathrm{i}}$ When system returns to its initial state, $\therefore \quad \Delta \mathrm{U}=0$
