272739
Calculate the change of entropy for the process, water (liquid) to water (vapour) involving $\Delta \mathrm{H}_{\text {rap }}=40850 \mathrm{~J} \mathrm{~mol}^{-1}$ at $373 \mathrm{~K}$.
272740
The temperature of $K$ at which $\Delta G=0$, for a given reaction with $\Delta \mathrm{H}=-20.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\Delta S$ $=-50.0 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ is
272741
What is the entropy change in $\mathrm{JK}^{-1}$ during the melting of $27.3 \mathrm{~g}$ of ice at $0^{\circ} \mathrm{C}$ ? (Latent heat of fusion of ice $=330 \mathrm{Jg}^{-1}$ )
1 330
2 12.1
3 3.3
4 33
Explanation:
We know that, $\Delta \mathrm{S}_{\mathrm{f}}=\frac{\Delta \mathrm{H}_{\mathrm{f}}}{\mathrm{T}_{\mathrm{f}}}$ Given that, Latent heat of fusion of ice $=330 \mathrm{Jg}^{-1}$ $\mathrm{T}=0^{\circ} \mathrm{C}=273 \mathrm{~K}$ So, $\Delta \mathrm{S}_{\mathrm{f}} =\frac{330 \times 27.3}{273}$ $=33 \mathrm{JK}^{-1}$
AP-EAMCET- (Engg.)-2011
Thermodynamics
272743
The enthalpy of vaporization of a liquid is 35.2 $\mathrm{kJ} \mathrm{mol}{ }^{-1}$ at $110.6^{\circ} \mathrm{C}$. The entropy change for the process will be
1 $9.18 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
2 $31.83 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
3 $91.76 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
4 $318.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
Explanation:
We know that, $\text { Entropy of vaporisation }=\frac{\text { Enthalpy of vaporisation }}{\text { Boiling point }}$ $(\Delta S)_{\operatorname{tap}}=\frac{(\Delta \mathrm{H})_{\operatorname{tap}}}{\mathrm{T}_{\mathrm{bp}}}$ $\text { or, Entropy of vaporisation }(\Delta S)=\frac{35.2}{110.6} \mathrm{Jmol}^{-1} \mathrm{~K}^{-1}$ $=0.3183 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}=318.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
AMU-2006
Thermodynamics
272744
Enthalpy change when lg water is frozen at $0^{\circ} \mathrm{C}$ is : $\left(\Delta \mathrm{H}_{\text {fus }}=1.435 \mathrm{kcal} \mathrm{mol}^{-1}\right)$
1 $0.0797 \mathrm{kcal}$
2 $-0.0797 \mathrm{kcal}$
3 $1.435 \mathrm{kcal}$
4 $-1.435 \mathrm{kcal}$
Explanation:
Given that, $(\Delta \mathrm{H})_{\text {firsion }}=1.435 \mathrm{kcal} / \mathrm{mol}$ - For 1 mole heat given $1.435 \mathrm{kcal}$ - For water $\frac{1}{18}$ mole $=1.435 \times \frac{1}{18} \mathrm{kcal}$ heat $=0.07972 \mathrm{kcal}$ Since heat will be released $\Rightarrow \Delta H=-0.07972 \mathrm{kcal}$
272739
Calculate the change of entropy for the process, water (liquid) to water (vapour) involving $\Delta \mathrm{H}_{\text {rap }}=40850 \mathrm{~J} \mathrm{~mol}^{-1}$ at $373 \mathrm{~K}$.
272740
The temperature of $K$ at which $\Delta G=0$, for a given reaction with $\Delta \mathrm{H}=-20.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\Delta S$ $=-50.0 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ is
272741
What is the entropy change in $\mathrm{JK}^{-1}$ during the melting of $27.3 \mathrm{~g}$ of ice at $0^{\circ} \mathrm{C}$ ? (Latent heat of fusion of ice $=330 \mathrm{Jg}^{-1}$ )
1 330
2 12.1
3 3.3
4 33
Explanation:
We know that, $\Delta \mathrm{S}_{\mathrm{f}}=\frac{\Delta \mathrm{H}_{\mathrm{f}}}{\mathrm{T}_{\mathrm{f}}}$ Given that, Latent heat of fusion of ice $=330 \mathrm{Jg}^{-1}$ $\mathrm{T}=0^{\circ} \mathrm{C}=273 \mathrm{~K}$ So, $\Delta \mathrm{S}_{\mathrm{f}} =\frac{330 \times 27.3}{273}$ $=33 \mathrm{JK}^{-1}$
AP-EAMCET- (Engg.)-2011
Thermodynamics
272743
The enthalpy of vaporization of a liquid is 35.2 $\mathrm{kJ} \mathrm{mol}{ }^{-1}$ at $110.6^{\circ} \mathrm{C}$. The entropy change for the process will be
1 $9.18 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
2 $31.83 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
3 $91.76 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
4 $318.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
Explanation:
We know that, $\text { Entropy of vaporisation }=\frac{\text { Enthalpy of vaporisation }}{\text { Boiling point }}$ $(\Delta S)_{\operatorname{tap}}=\frac{(\Delta \mathrm{H})_{\operatorname{tap}}}{\mathrm{T}_{\mathrm{bp}}}$ $\text { or, Entropy of vaporisation }(\Delta S)=\frac{35.2}{110.6} \mathrm{Jmol}^{-1} \mathrm{~K}^{-1}$ $=0.3183 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}=318.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
AMU-2006
Thermodynamics
272744
Enthalpy change when lg water is frozen at $0^{\circ} \mathrm{C}$ is : $\left(\Delta \mathrm{H}_{\text {fus }}=1.435 \mathrm{kcal} \mathrm{mol}^{-1}\right)$
1 $0.0797 \mathrm{kcal}$
2 $-0.0797 \mathrm{kcal}$
3 $1.435 \mathrm{kcal}$
4 $-1.435 \mathrm{kcal}$
Explanation:
Given that, $(\Delta \mathrm{H})_{\text {firsion }}=1.435 \mathrm{kcal} / \mathrm{mol}$ - For 1 mole heat given $1.435 \mathrm{kcal}$ - For water $\frac{1}{18}$ mole $=1.435 \times \frac{1}{18} \mathrm{kcal}$ heat $=0.07972 \mathrm{kcal}$ Since heat will be released $\Rightarrow \Delta H=-0.07972 \mathrm{kcal}$
272739
Calculate the change of entropy for the process, water (liquid) to water (vapour) involving $\Delta \mathrm{H}_{\text {rap }}=40850 \mathrm{~J} \mathrm{~mol}^{-1}$ at $373 \mathrm{~K}$.
272740
The temperature of $K$ at which $\Delta G=0$, for a given reaction with $\Delta \mathrm{H}=-20.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\Delta S$ $=-50.0 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ is
272741
What is the entropy change in $\mathrm{JK}^{-1}$ during the melting of $27.3 \mathrm{~g}$ of ice at $0^{\circ} \mathrm{C}$ ? (Latent heat of fusion of ice $=330 \mathrm{Jg}^{-1}$ )
1 330
2 12.1
3 3.3
4 33
Explanation:
We know that, $\Delta \mathrm{S}_{\mathrm{f}}=\frac{\Delta \mathrm{H}_{\mathrm{f}}}{\mathrm{T}_{\mathrm{f}}}$ Given that, Latent heat of fusion of ice $=330 \mathrm{Jg}^{-1}$ $\mathrm{T}=0^{\circ} \mathrm{C}=273 \mathrm{~K}$ So, $\Delta \mathrm{S}_{\mathrm{f}} =\frac{330 \times 27.3}{273}$ $=33 \mathrm{JK}^{-1}$
AP-EAMCET- (Engg.)-2011
Thermodynamics
272743
The enthalpy of vaporization of a liquid is 35.2 $\mathrm{kJ} \mathrm{mol}{ }^{-1}$ at $110.6^{\circ} \mathrm{C}$. The entropy change for the process will be
1 $9.18 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
2 $31.83 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
3 $91.76 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
4 $318.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
Explanation:
We know that, $\text { Entropy of vaporisation }=\frac{\text { Enthalpy of vaporisation }}{\text { Boiling point }}$ $(\Delta S)_{\operatorname{tap}}=\frac{(\Delta \mathrm{H})_{\operatorname{tap}}}{\mathrm{T}_{\mathrm{bp}}}$ $\text { or, Entropy of vaporisation }(\Delta S)=\frac{35.2}{110.6} \mathrm{Jmol}^{-1} \mathrm{~K}^{-1}$ $=0.3183 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}=318.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
AMU-2006
Thermodynamics
272744
Enthalpy change when lg water is frozen at $0^{\circ} \mathrm{C}$ is : $\left(\Delta \mathrm{H}_{\text {fus }}=1.435 \mathrm{kcal} \mathrm{mol}^{-1}\right)$
1 $0.0797 \mathrm{kcal}$
2 $-0.0797 \mathrm{kcal}$
3 $1.435 \mathrm{kcal}$
4 $-1.435 \mathrm{kcal}$
Explanation:
Given that, $(\Delta \mathrm{H})_{\text {firsion }}=1.435 \mathrm{kcal} / \mathrm{mol}$ - For 1 mole heat given $1.435 \mathrm{kcal}$ - For water $\frac{1}{18}$ mole $=1.435 \times \frac{1}{18} \mathrm{kcal}$ heat $=0.07972 \mathrm{kcal}$ Since heat will be released $\Rightarrow \Delta H=-0.07972 \mathrm{kcal}$
272739
Calculate the change of entropy for the process, water (liquid) to water (vapour) involving $\Delta \mathrm{H}_{\text {rap }}=40850 \mathrm{~J} \mathrm{~mol}^{-1}$ at $373 \mathrm{~K}$.
272740
The temperature of $K$ at which $\Delta G=0$, for a given reaction with $\Delta \mathrm{H}=-20.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\Delta S$ $=-50.0 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ is
272741
What is the entropy change in $\mathrm{JK}^{-1}$ during the melting of $27.3 \mathrm{~g}$ of ice at $0^{\circ} \mathrm{C}$ ? (Latent heat of fusion of ice $=330 \mathrm{Jg}^{-1}$ )
1 330
2 12.1
3 3.3
4 33
Explanation:
We know that, $\Delta \mathrm{S}_{\mathrm{f}}=\frac{\Delta \mathrm{H}_{\mathrm{f}}}{\mathrm{T}_{\mathrm{f}}}$ Given that, Latent heat of fusion of ice $=330 \mathrm{Jg}^{-1}$ $\mathrm{T}=0^{\circ} \mathrm{C}=273 \mathrm{~K}$ So, $\Delta \mathrm{S}_{\mathrm{f}} =\frac{330 \times 27.3}{273}$ $=33 \mathrm{JK}^{-1}$
AP-EAMCET- (Engg.)-2011
Thermodynamics
272743
The enthalpy of vaporization of a liquid is 35.2 $\mathrm{kJ} \mathrm{mol}{ }^{-1}$ at $110.6^{\circ} \mathrm{C}$. The entropy change for the process will be
1 $9.18 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
2 $31.83 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
3 $91.76 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
4 $318.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
Explanation:
We know that, $\text { Entropy of vaporisation }=\frac{\text { Enthalpy of vaporisation }}{\text { Boiling point }}$ $(\Delta S)_{\operatorname{tap}}=\frac{(\Delta \mathrm{H})_{\operatorname{tap}}}{\mathrm{T}_{\mathrm{bp}}}$ $\text { or, Entropy of vaporisation }(\Delta S)=\frac{35.2}{110.6} \mathrm{Jmol}^{-1} \mathrm{~K}^{-1}$ $=0.3183 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}=318.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
AMU-2006
Thermodynamics
272744
Enthalpy change when lg water is frozen at $0^{\circ} \mathrm{C}$ is : $\left(\Delta \mathrm{H}_{\text {fus }}=1.435 \mathrm{kcal} \mathrm{mol}^{-1}\right)$
1 $0.0797 \mathrm{kcal}$
2 $-0.0797 \mathrm{kcal}$
3 $1.435 \mathrm{kcal}$
4 $-1.435 \mathrm{kcal}$
Explanation:
Given that, $(\Delta \mathrm{H})_{\text {firsion }}=1.435 \mathrm{kcal} / \mathrm{mol}$ - For 1 mole heat given $1.435 \mathrm{kcal}$ - For water $\frac{1}{18}$ mole $=1.435 \times \frac{1}{18} \mathrm{kcal}$ heat $=0.07972 \mathrm{kcal}$ Since heat will be released $\Rightarrow \Delta H=-0.07972 \mathrm{kcal}$
272739
Calculate the change of entropy for the process, water (liquid) to water (vapour) involving $\Delta \mathrm{H}_{\text {rap }}=40850 \mathrm{~J} \mathrm{~mol}^{-1}$ at $373 \mathrm{~K}$.
272740
The temperature of $K$ at which $\Delta G=0$, for a given reaction with $\Delta \mathrm{H}=-20.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\Delta S$ $=-50.0 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ is
272741
What is the entropy change in $\mathrm{JK}^{-1}$ during the melting of $27.3 \mathrm{~g}$ of ice at $0^{\circ} \mathrm{C}$ ? (Latent heat of fusion of ice $=330 \mathrm{Jg}^{-1}$ )
1 330
2 12.1
3 3.3
4 33
Explanation:
We know that, $\Delta \mathrm{S}_{\mathrm{f}}=\frac{\Delta \mathrm{H}_{\mathrm{f}}}{\mathrm{T}_{\mathrm{f}}}$ Given that, Latent heat of fusion of ice $=330 \mathrm{Jg}^{-1}$ $\mathrm{T}=0^{\circ} \mathrm{C}=273 \mathrm{~K}$ So, $\Delta \mathrm{S}_{\mathrm{f}} =\frac{330 \times 27.3}{273}$ $=33 \mathrm{JK}^{-1}$
AP-EAMCET- (Engg.)-2011
Thermodynamics
272743
The enthalpy of vaporization of a liquid is 35.2 $\mathrm{kJ} \mathrm{mol}{ }^{-1}$ at $110.6^{\circ} \mathrm{C}$. The entropy change for the process will be
1 $9.18 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
2 $31.83 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
3 $91.76 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
4 $318.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$
Explanation:
We know that, $\text { Entropy of vaporisation }=\frac{\text { Enthalpy of vaporisation }}{\text { Boiling point }}$ $(\Delta S)_{\operatorname{tap}}=\frac{(\Delta \mathrm{H})_{\operatorname{tap}}}{\mathrm{T}_{\mathrm{bp}}}$ $\text { or, Entropy of vaporisation }(\Delta S)=\frac{35.2}{110.6} \mathrm{Jmol}^{-1} \mathrm{~K}^{-1}$ $=0.3183 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}=318.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
AMU-2006
Thermodynamics
272744
Enthalpy change when lg water is frozen at $0^{\circ} \mathrm{C}$ is : $\left(\Delta \mathrm{H}_{\text {fus }}=1.435 \mathrm{kcal} \mathrm{mol}^{-1}\right)$
1 $0.0797 \mathrm{kcal}$
2 $-0.0797 \mathrm{kcal}$
3 $1.435 \mathrm{kcal}$
4 $-1.435 \mathrm{kcal}$
Explanation:
Given that, $(\Delta \mathrm{H})_{\text {firsion }}=1.435 \mathrm{kcal} / \mathrm{mol}$ - For 1 mole heat given $1.435 \mathrm{kcal}$ - For water $\frac{1}{18}$ mole $=1.435 \times \frac{1}{18} \mathrm{kcal}$ heat $=0.07972 \mathrm{kcal}$ Since heat will be released $\Rightarrow \Delta H=-0.07972 \mathrm{kcal}$
