The higher is the pressure lower become entropy. When the pressure increase on the gases molecule kinetic energy of the particle decrease with respect to the pressure. This is due to the particle comes closer and intermolecular distance decreases the entropy. Entropy is a state function that is often referred to as the state of disorder of a system. - $\mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}(\mathrm{g})$ entropy increases because no. of atoms increases. - $\mathrm{N}_2(\mathrm{~g}, 1 \mathrm{~atm}) \rightarrow \mathrm{N}_2(\mathrm{~g}, 8 \mathrm{~atm})$ entropy decreases as pressure increase and randomness decreases. - $2 \mathrm{SO}_3(\mathrm{~g}) \rightarrow 2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g})$ entropy increases because no. of atoms increases. - $\mathrm{C}_{(\text {dimond) }} \rightarrow \mathrm{C}_{\text {(graphite) }}$ entropy increases as the compactness of system decreases and randomness increases.
BITSAT 2008
Thermodynamics
272746
For a given reaction, $\Delta \mathrm{H}=35.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\Delta S=83.6 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$. The reaction is spontaneous at : (Assume that $\Delta H$ and $\Delta S$ do not vary with temperature)
1 $\mathrm{T}>425 \mathrm{~K}$
2 All temperature
3 $\mathrm{T}>298 \mathrm{~K}$
4 $\mathrm{T}<425 \mathrm{~K}$
Explanation:
Given that $\Delta \mathrm{H}=35.5 \mathrm{~kJ} / \mathrm{mol}=35500 \mathrm{~J} / \mathrm{mol}$ $\Delta \mathrm{S}=83.6 \mathrm{~J} / \mathrm{K} / \mathrm{mol}$ $\mathrm{We} \mathrm{know}$ that, $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ For spontaneous $\Delta \mathrm{G}<0$ $\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}<0$ $\mathrm{T}>\frac{\Delta \mathrm{H}}{\Delta \mathrm{S}}$ $\mathrm{T}>\frac{35500}{83.6}$ $\mathrm{T}>425 \mathrm{~K}$ So, the given reaction will be spontaneous at $\mathrm{T}>425 \mathrm{~K}$.
BITSAT 2017
Thermodynamics
272754
The values of $\Delta H$ and $\Delta \mathrm{S}$ for a reaction are respectively $30 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $100 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$. Then the temperature above which the reaction will become spontaneous is
272760
When the same quantity of heat is absorbed by a system at two different temperatures $T_1$ and $T_2$, such that $T_1>T_2$, change in entropies are $\Delta S_1$ and $\Delta S_2$ respectively. Then
1 $\Delta \mathrm{S}_1<\Delta \mathrm{S}_2$
2 $\Delta \mathrm{S}_1=\Delta \mathrm{S}_2$
3 $\mathrm{S}_1>\mathrm{S}_2$
4 $\Delta \mathrm{S}_2<\Delta \mathrm{S}_1$
Explanation:
Entropy change occur due to transfer of heat. $\Delta \mathrm{S}=\mathrm{q} / \mathrm{T} \quad[\mathrm{q}=$ constant $]$ So, $\quad \Delta \mathrm{S} \propto \frac{1}{\mathrm{~T}}$ Given, $\mathrm{T}_1>\mathrm{T}_2$ So, actual relation between the entropies will be $\left(\Delta \mathrm{S}_2>\Delta \mathrm{S}_1\right)$
The higher is the pressure lower become entropy. When the pressure increase on the gases molecule kinetic energy of the particle decrease with respect to the pressure. This is due to the particle comes closer and intermolecular distance decreases the entropy. Entropy is a state function that is often referred to as the state of disorder of a system. - $\mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}(\mathrm{g})$ entropy increases because no. of atoms increases. - $\mathrm{N}_2(\mathrm{~g}, 1 \mathrm{~atm}) \rightarrow \mathrm{N}_2(\mathrm{~g}, 8 \mathrm{~atm})$ entropy decreases as pressure increase and randomness decreases. - $2 \mathrm{SO}_3(\mathrm{~g}) \rightarrow 2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g})$ entropy increases because no. of atoms increases. - $\mathrm{C}_{(\text {dimond) }} \rightarrow \mathrm{C}_{\text {(graphite) }}$ entropy increases as the compactness of system decreases and randomness increases.
BITSAT 2008
Thermodynamics
272746
For a given reaction, $\Delta \mathrm{H}=35.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\Delta S=83.6 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$. The reaction is spontaneous at : (Assume that $\Delta H$ and $\Delta S$ do not vary with temperature)
1 $\mathrm{T}>425 \mathrm{~K}$
2 All temperature
3 $\mathrm{T}>298 \mathrm{~K}$
4 $\mathrm{T}<425 \mathrm{~K}$
Explanation:
Given that $\Delta \mathrm{H}=35.5 \mathrm{~kJ} / \mathrm{mol}=35500 \mathrm{~J} / \mathrm{mol}$ $\Delta \mathrm{S}=83.6 \mathrm{~J} / \mathrm{K} / \mathrm{mol}$ $\mathrm{We} \mathrm{know}$ that, $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ For spontaneous $\Delta \mathrm{G}<0$ $\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}<0$ $\mathrm{T}>\frac{\Delta \mathrm{H}}{\Delta \mathrm{S}}$ $\mathrm{T}>\frac{35500}{83.6}$ $\mathrm{T}>425 \mathrm{~K}$ So, the given reaction will be spontaneous at $\mathrm{T}>425 \mathrm{~K}$.
BITSAT 2017
Thermodynamics
272754
The values of $\Delta H$ and $\Delta \mathrm{S}$ for a reaction are respectively $30 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $100 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$. Then the temperature above which the reaction will become spontaneous is
272760
When the same quantity of heat is absorbed by a system at two different temperatures $T_1$ and $T_2$, such that $T_1>T_2$, change in entropies are $\Delta S_1$ and $\Delta S_2$ respectively. Then
1 $\Delta \mathrm{S}_1<\Delta \mathrm{S}_2$
2 $\Delta \mathrm{S}_1=\Delta \mathrm{S}_2$
3 $\mathrm{S}_1>\mathrm{S}_2$
4 $\Delta \mathrm{S}_2<\Delta \mathrm{S}_1$
Explanation:
Entropy change occur due to transfer of heat. $\Delta \mathrm{S}=\mathrm{q} / \mathrm{T} \quad[\mathrm{q}=$ constant $]$ So, $\quad \Delta \mathrm{S} \propto \frac{1}{\mathrm{~T}}$ Given, $\mathrm{T}_1>\mathrm{T}_2$ So, actual relation between the entropies will be $\left(\Delta \mathrm{S}_2>\Delta \mathrm{S}_1\right)$
The higher is the pressure lower become entropy. When the pressure increase on the gases molecule kinetic energy of the particle decrease with respect to the pressure. This is due to the particle comes closer and intermolecular distance decreases the entropy. Entropy is a state function that is often referred to as the state of disorder of a system. - $\mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}(\mathrm{g})$ entropy increases because no. of atoms increases. - $\mathrm{N}_2(\mathrm{~g}, 1 \mathrm{~atm}) \rightarrow \mathrm{N}_2(\mathrm{~g}, 8 \mathrm{~atm})$ entropy decreases as pressure increase and randomness decreases. - $2 \mathrm{SO}_3(\mathrm{~g}) \rightarrow 2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g})$ entropy increases because no. of atoms increases. - $\mathrm{C}_{(\text {dimond) }} \rightarrow \mathrm{C}_{\text {(graphite) }}$ entropy increases as the compactness of system decreases and randomness increases.
BITSAT 2008
Thermodynamics
272746
For a given reaction, $\Delta \mathrm{H}=35.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\Delta S=83.6 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$. The reaction is spontaneous at : (Assume that $\Delta H$ and $\Delta S$ do not vary with temperature)
1 $\mathrm{T}>425 \mathrm{~K}$
2 All temperature
3 $\mathrm{T}>298 \mathrm{~K}$
4 $\mathrm{T}<425 \mathrm{~K}$
Explanation:
Given that $\Delta \mathrm{H}=35.5 \mathrm{~kJ} / \mathrm{mol}=35500 \mathrm{~J} / \mathrm{mol}$ $\Delta \mathrm{S}=83.6 \mathrm{~J} / \mathrm{K} / \mathrm{mol}$ $\mathrm{We} \mathrm{know}$ that, $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ For spontaneous $\Delta \mathrm{G}<0$ $\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}<0$ $\mathrm{T}>\frac{\Delta \mathrm{H}}{\Delta \mathrm{S}}$ $\mathrm{T}>\frac{35500}{83.6}$ $\mathrm{T}>425 \mathrm{~K}$ So, the given reaction will be spontaneous at $\mathrm{T}>425 \mathrm{~K}$.
BITSAT 2017
Thermodynamics
272754
The values of $\Delta H$ and $\Delta \mathrm{S}$ for a reaction are respectively $30 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $100 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$. Then the temperature above which the reaction will become spontaneous is
272760
When the same quantity of heat is absorbed by a system at two different temperatures $T_1$ and $T_2$, such that $T_1>T_2$, change in entropies are $\Delta S_1$ and $\Delta S_2$ respectively. Then
1 $\Delta \mathrm{S}_1<\Delta \mathrm{S}_2$
2 $\Delta \mathrm{S}_1=\Delta \mathrm{S}_2$
3 $\mathrm{S}_1>\mathrm{S}_2$
4 $\Delta \mathrm{S}_2<\Delta \mathrm{S}_1$
Explanation:
Entropy change occur due to transfer of heat. $\Delta \mathrm{S}=\mathrm{q} / \mathrm{T} \quad[\mathrm{q}=$ constant $]$ So, $\quad \Delta \mathrm{S} \propto \frac{1}{\mathrm{~T}}$ Given, $\mathrm{T}_1>\mathrm{T}_2$ So, actual relation between the entropies will be $\left(\Delta \mathrm{S}_2>\Delta \mathrm{S}_1\right)$
The higher is the pressure lower become entropy. When the pressure increase on the gases molecule kinetic energy of the particle decrease with respect to the pressure. This is due to the particle comes closer and intermolecular distance decreases the entropy. Entropy is a state function that is often referred to as the state of disorder of a system. - $\mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{H}(\mathrm{g})$ entropy increases because no. of atoms increases. - $\mathrm{N}_2(\mathrm{~g}, 1 \mathrm{~atm}) \rightarrow \mathrm{N}_2(\mathrm{~g}, 8 \mathrm{~atm})$ entropy decreases as pressure increase and randomness decreases. - $2 \mathrm{SO}_3(\mathrm{~g}) \rightarrow 2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g})$ entropy increases because no. of atoms increases. - $\mathrm{C}_{(\text {dimond) }} \rightarrow \mathrm{C}_{\text {(graphite) }}$ entropy increases as the compactness of system decreases and randomness increases.
BITSAT 2008
Thermodynamics
272746
For a given reaction, $\Delta \mathrm{H}=35.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\Delta S=83.6 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$. The reaction is spontaneous at : (Assume that $\Delta H$ and $\Delta S$ do not vary with temperature)
1 $\mathrm{T}>425 \mathrm{~K}$
2 All temperature
3 $\mathrm{T}>298 \mathrm{~K}$
4 $\mathrm{T}<425 \mathrm{~K}$
Explanation:
Given that $\Delta \mathrm{H}=35.5 \mathrm{~kJ} / \mathrm{mol}=35500 \mathrm{~J} / \mathrm{mol}$ $\Delta \mathrm{S}=83.6 \mathrm{~J} / \mathrm{K} / \mathrm{mol}$ $\mathrm{We} \mathrm{know}$ that, $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$ For spontaneous $\Delta \mathrm{G}<0$ $\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}<0$ $\mathrm{T}>\frac{\Delta \mathrm{H}}{\Delta \mathrm{S}}$ $\mathrm{T}>\frac{35500}{83.6}$ $\mathrm{T}>425 \mathrm{~K}$ So, the given reaction will be spontaneous at $\mathrm{T}>425 \mathrm{~K}$.
BITSAT 2017
Thermodynamics
272754
The values of $\Delta H$ and $\Delta \mathrm{S}$ for a reaction are respectively $30 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $100 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$. Then the temperature above which the reaction will become spontaneous is
272760
When the same quantity of heat is absorbed by a system at two different temperatures $T_1$ and $T_2$, such that $T_1>T_2$, change in entropies are $\Delta S_1$ and $\Delta S_2$ respectively. Then
1 $\Delta \mathrm{S}_1<\Delta \mathrm{S}_2$
2 $\Delta \mathrm{S}_1=\Delta \mathrm{S}_2$
3 $\mathrm{S}_1>\mathrm{S}_2$
4 $\Delta \mathrm{S}_2<\Delta \mathrm{S}_1$
Explanation:
Entropy change occur due to transfer of heat. $\Delta \mathrm{S}=\mathrm{q} / \mathrm{T} \quad[\mathrm{q}=$ constant $]$ So, $\quad \Delta \mathrm{S} \propto \frac{1}{\mathrm{~T}}$ Given, $\mathrm{T}_1>\mathrm{T}_2$ So, actual relation between the entropies will be $\left(\Delta \mathrm{S}_2>\Delta \mathrm{S}_1\right)$