272404
Standard eptropy of $X_2, Y_2$ and $X Y_3$ are 60,40 and $50 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$, respectively. For the reaction, $\frac{1}{2} \mathrm{X}_2+\frac{3}{2} Y_2$ ? $\mathrm{XY}_3, \Delta \mathrm{H}=-30 \mathrm{~kJ}$, to be at equilibrium, the temperature will be
1 $1250 \mathrm{~K}$
2 $500 \mathrm{~K}$
3 $750 \mathrm{~K}$
4 $1000 \mathrm{~K}$
Explanation:
For the reaction $\frac{1}{2} \mathrm{X}_2+\frac{3}{2} \mathrm{Y}_2 \rightarrow \mathrm{XY}_3 ; \Delta \mathrm{H}=-30 \mathrm{~kJ} \text { (given) }$ Calculating the $\Delta S$ for the above reaction- $\Delta \mathrm{S}=50-\left[\frac{1}{2} \times 60+\frac{3}{2} \times 40\right] \mathrm{JK}^{-1}$ $\Delta \mathrm{S}=50-[30+60]$ $\Delta \mathrm{S}=-40 \mathrm{JK}^{-1}$ At equilibrium, $\Delta \mathrm{G}=0, \Delta \mathrm{G}=\Delta \mathrm{H}-\Delta \mathrm{ST}$ $\mathrm{T} \Delta \mathrm{S}=\Delta \mathrm{H}$ $\mathrm{T}=\frac{\Delta \mathrm{H}}{\Delta \mathrm{S}}=\frac{-30 \times 1000}{-40}$ $\mathrm{~T}=750 \mathrm{~K} .$
AIEEE-2008
Thermodynamics
272405
One mole of an ideal gas at $300 \mathrm{~K}$ is expanded isothermally from an initial volume of 1 litre to 10 litres. Then $\Delta S$ (cal deg $^{-1} \mathrm{~mol}^{-1}$ ) for this process is: $\left(R=2 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right)$
272407
For the reaction, $\mathrm{C}_3 \mathrm{H}_8(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 3 \mathrm{CO}_2(\mathrm{~g})+4 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$ at constant temperature, $\Delta H-\Delta \mathrm{E}$ is
1 $\mathrm{RT}$
2 $-3 \mathrm{RT}$
3 $3 \mathrm{RT}$
4 - RT
Explanation:
: For the reaction- $\mathrm{C}_3 \mathrm{H}_8(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \rightarrow 3 \mathrm{CO}_2(\mathrm{~g})+4 \mathrm{H}_2 \mathrm{O}(l)$ Given that, at constant temperature Now, from the following equation- $\Delta \mathrm{H}=\Delta \mathrm{E}+\mathrm{P} \Delta \mathrm{V}\left(\because \mathrm{P} \Delta \mathrm{V}=\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}\right)$ or $\Delta \mathrm{H}=\Delta \mathrm{E}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}$ Where, $\Delta \mathrm{n}_{\mathrm{g}}=\mathrm{n}_{\mathrm{p}}-\mathrm{n}_{\mathrm{r}}=3-6=-3$ $\therefore \quad \Delta \mathrm{H}-\Delta \mathrm{E}=-3 \mathrm{RT}$
VITEEE- 2011
Thermodynamics
272409
Calculate $\Delta \mathrm{H}^{\circ}$ for the reaction, $\mathrm{Na}_2 \mathrm{O}(\mathrm{s})+\mathrm{SO}_3(\mathrm{~g}) \rightarrow \mathrm{NaSO}_4(\mathrm{~g})$ given the following :
NEET Test Series from KOTA - 10 Papers In MS WORD
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Thermodynamics
272404
Standard eptropy of $X_2, Y_2$ and $X Y_3$ are 60,40 and $50 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$, respectively. For the reaction, $\frac{1}{2} \mathrm{X}_2+\frac{3}{2} Y_2$ ? $\mathrm{XY}_3, \Delta \mathrm{H}=-30 \mathrm{~kJ}$, to be at equilibrium, the temperature will be
1 $1250 \mathrm{~K}$
2 $500 \mathrm{~K}$
3 $750 \mathrm{~K}$
4 $1000 \mathrm{~K}$
Explanation:
For the reaction $\frac{1}{2} \mathrm{X}_2+\frac{3}{2} \mathrm{Y}_2 \rightarrow \mathrm{XY}_3 ; \Delta \mathrm{H}=-30 \mathrm{~kJ} \text { (given) }$ Calculating the $\Delta S$ for the above reaction- $\Delta \mathrm{S}=50-\left[\frac{1}{2} \times 60+\frac{3}{2} \times 40\right] \mathrm{JK}^{-1}$ $\Delta \mathrm{S}=50-[30+60]$ $\Delta \mathrm{S}=-40 \mathrm{JK}^{-1}$ At equilibrium, $\Delta \mathrm{G}=0, \Delta \mathrm{G}=\Delta \mathrm{H}-\Delta \mathrm{ST}$ $\mathrm{T} \Delta \mathrm{S}=\Delta \mathrm{H}$ $\mathrm{T}=\frac{\Delta \mathrm{H}}{\Delta \mathrm{S}}=\frac{-30 \times 1000}{-40}$ $\mathrm{~T}=750 \mathrm{~K} .$
AIEEE-2008
Thermodynamics
272405
One mole of an ideal gas at $300 \mathrm{~K}$ is expanded isothermally from an initial volume of 1 litre to 10 litres. Then $\Delta S$ (cal deg $^{-1} \mathrm{~mol}^{-1}$ ) for this process is: $\left(R=2 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right)$
272407
For the reaction, $\mathrm{C}_3 \mathrm{H}_8(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 3 \mathrm{CO}_2(\mathrm{~g})+4 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$ at constant temperature, $\Delta H-\Delta \mathrm{E}$ is
1 $\mathrm{RT}$
2 $-3 \mathrm{RT}$
3 $3 \mathrm{RT}$
4 - RT
Explanation:
: For the reaction- $\mathrm{C}_3 \mathrm{H}_8(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \rightarrow 3 \mathrm{CO}_2(\mathrm{~g})+4 \mathrm{H}_2 \mathrm{O}(l)$ Given that, at constant temperature Now, from the following equation- $\Delta \mathrm{H}=\Delta \mathrm{E}+\mathrm{P} \Delta \mathrm{V}\left(\because \mathrm{P} \Delta \mathrm{V}=\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}\right)$ or $\Delta \mathrm{H}=\Delta \mathrm{E}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}$ Where, $\Delta \mathrm{n}_{\mathrm{g}}=\mathrm{n}_{\mathrm{p}}-\mathrm{n}_{\mathrm{r}}=3-6=-3$ $\therefore \quad \Delta \mathrm{H}-\Delta \mathrm{E}=-3 \mathrm{RT}$
VITEEE- 2011
Thermodynamics
272409
Calculate $\Delta \mathrm{H}^{\circ}$ for the reaction, $\mathrm{Na}_2 \mathrm{O}(\mathrm{s})+\mathrm{SO}_3(\mathrm{~g}) \rightarrow \mathrm{NaSO}_4(\mathrm{~g})$ given the following :
272404
Standard eptropy of $X_2, Y_2$ and $X Y_3$ are 60,40 and $50 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$, respectively. For the reaction, $\frac{1}{2} \mathrm{X}_2+\frac{3}{2} Y_2$ ? $\mathrm{XY}_3, \Delta \mathrm{H}=-30 \mathrm{~kJ}$, to be at equilibrium, the temperature will be
1 $1250 \mathrm{~K}$
2 $500 \mathrm{~K}$
3 $750 \mathrm{~K}$
4 $1000 \mathrm{~K}$
Explanation:
For the reaction $\frac{1}{2} \mathrm{X}_2+\frac{3}{2} \mathrm{Y}_2 \rightarrow \mathrm{XY}_3 ; \Delta \mathrm{H}=-30 \mathrm{~kJ} \text { (given) }$ Calculating the $\Delta S$ for the above reaction- $\Delta \mathrm{S}=50-\left[\frac{1}{2} \times 60+\frac{3}{2} \times 40\right] \mathrm{JK}^{-1}$ $\Delta \mathrm{S}=50-[30+60]$ $\Delta \mathrm{S}=-40 \mathrm{JK}^{-1}$ At equilibrium, $\Delta \mathrm{G}=0, \Delta \mathrm{G}=\Delta \mathrm{H}-\Delta \mathrm{ST}$ $\mathrm{T} \Delta \mathrm{S}=\Delta \mathrm{H}$ $\mathrm{T}=\frac{\Delta \mathrm{H}}{\Delta \mathrm{S}}=\frac{-30 \times 1000}{-40}$ $\mathrm{~T}=750 \mathrm{~K} .$
AIEEE-2008
Thermodynamics
272405
One mole of an ideal gas at $300 \mathrm{~K}$ is expanded isothermally from an initial volume of 1 litre to 10 litres. Then $\Delta S$ (cal deg $^{-1} \mathrm{~mol}^{-1}$ ) for this process is: $\left(R=2 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right)$
272407
For the reaction, $\mathrm{C}_3 \mathrm{H}_8(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 3 \mathrm{CO}_2(\mathrm{~g})+4 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$ at constant temperature, $\Delta H-\Delta \mathrm{E}$ is
1 $\mathrm{RT}$
2 $-3 \mathrm{RT}$
3 $3 \mathrm{RT}$
4 - RT
Explanation:
: For the reaction- $\mathrm{C}_3 \mathrm{H}_8(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \rightarrow 3 \mathrm{CO}_2(\mathrm{~g})+4 \mathrm{H}_2 \mathrm{O}(l)$ Given that, at constant temperature Now, from the following equation- $\Delta \mathrm{H}=\Delta \mathrm{E}+\mathrm{P} \Delta \mathrm{V}\left(\because \mathrm{P} \Delta \mathrm{V}=\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}\right)$ or $\Delta \mathrm{H}=\Delta \mathrm{E}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}$ Where, $\Delta \mathrm{n}_{\mathrm{g}}=\mathrm{n}_{\mathrm{p}}-\mathrm{n}_{\mathrm{r}}=3-6=-3$ $\therefore \quad \Delta \mathrm{H}-\Delta \mathrm{E}=-3 \mathrm{RT}$
VITEEE- 2011
Thermodynamics
272409
Calculate $\Delta \mathrm{H}^{\circ}$ for the reaction, $\mathrm{Na}_2 \mathrm{O}(\mathrm{s})+\mathrm{SO}_3(\mathrm{~g}) \rightarrow \mathrm{NaSO}_4(\mathrm{~g})$ given the following :
272404
Standard eptropy of $X_2, Y_2$ and $X Y_3$ are 60,40 and $50 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$, respectively. For the reaction, $\frac{1}{2} \mathrm{X}_2+\frac{3}{2} Y_2$ ? $\mathrm{XY}_3, \Delta \mathrm{H}=-30 \mathrm{~kJ}$, to be at equilibrium, the temperature will be
1 $1250 \mathrm{~K}$
2 $500 \mathrm{~K}$
3 $750 \mathrm{~K}$
4 $1000 \mathrm{~K}$
Explanation:
For the reaction $\frac{1}{2} \mathrm{X}_2+\frac{3}{2} \mathrm{Y}_2 \rightarrow \mathrm{XY}_3 ; \Delta \mathrm{H}=-30 \mathrm{~kJ} \text { (given) }$ Calculating the $\Delta S$ for the above reaction- $\Delta \mathrm{S}=50-\left[\frac{1}{2} \times 60+\frac{3}{2} \times 40\right] \mathrm{JK}^{-1}$ $\Delta \mathrm{S}=50-[30+60]$ $\Delta \mathrm{S}=-40 \mathrm{JK}^{-1}$ At equilibrium, $\Delta \mathrm{G}=0, \Delta \mathrm{G}=\Delta \mathrm{H}-\Delta \mathrm{ST}$ $\mathrm{T} \Delta \mathrm{S}=\Delta \mathrm{H}$ $\mathrm{T}=\frac{\Delta \mathrm{H}}{\Delta \mathrm{S}}=\frac{-30 \times 1000}{-40}$ $\mathrm{~T}=750 \mathrm{~K} .$
AIEEE-2008
Thermodynamics
272405
One mole of an ideal gas at $300 \mathrm{~K}$ is expanded isothermally from an initial volume of 1 litre to 10 litres. Then $\Delta S$ (cal deg $^{-1} \mathrm{~mol}^{-1}$ ) for this process is: $\left(R=2 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right)$
272407
For the reaction, $\mathrm{C}_3 \mathrm{H}_8(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 3 \mathrm{CO}_2(\mathrm{~g})+4 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$ at constant temperature, $\Delta H-\Delta \mathrm{E}$ is
1 $\mathrm{RT}$
2 $-3 \mathrm{RT}$
3 $3 \mathrm{RT}$
4 - RT
Explanation:
: For the reaction- $\mathrm{C}_3 \mathrm{H}_8(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \rightarrow 3 \mathrm{CO}_2(\mathrm{~g})+4 \mathrm{H}_2 \mathrm{O}(l)$ Given that, at constant temperature Now, from the following equation- $\Delta \mathrm{H}=\Delta \mathrm{E}+\mathrm{P} \Delta \mathrm{V}\left(\because \mathrm{P} \Delta \mathrm{V}=\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}\right)$ or $\Delta \mathrm{H}=\Delta \mathrm{E}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}$ Where, $\Delta \mathrm{n}_{\mathrm{g}}=\mathrm{n}_{\mathrm{p}}-\mathrm{n}_{\mathrm{r}}=3-6=-3$ $\therefore \quad \Delta \mathrm{H}-\Delta \mathrm{E}=-3 \mathrm{RT}$
VITEEE- 2011
Thermodynamics
272409
Calculate $\Delta \mathrm{H}^{\circ}$ for the reaction, $\mathrm{Na}_2 \mathrm{O}(\mathrm{s})+\mathrm{SO}_3(\mathrm{~g}) \rightarrow \mathrm{NaSO}_4(\mathrm{~g})$ given the following :