273036
The energy that opposes dissolution of a solvent is
1 hydration energy
2 lattice energy
3 internal energy
4 bond energy
Explanation:
Lattice energy opposes dissolution of a solvent. Lattice energy is the amount of energy required to dissociate 1 mole of ionic crystals into ions. If hydration energy of solid is more than lattice energy, then solid dissolves.
BITSAT 2006
Thermodynamics
273035
The following is endothermic reaction
1 Decomposition of water
2 Conversion of graphite to diamond
3 Dehydrogenation of ethane to ethylene
4 All of the above
Explanation:
The reaction during which heat is observed known as endothermic reaction. (A) Combustion is always an exothermic procure. (B) Decomposition of $\mathrm{H}_2 \mathrm{O}$ requires energy (C) Dehydrogenation is an endothermic reaction. (D) Graphite is more stable than diamond and energy in needed to convert graphite into diamond.
BCECE-2010
Thermodynamics
273015
Bond enthalpies of $A_2, B_2$ and $A B$ are in the ratio $2: 1: 2$. If bond enthalpy of formation of $A B$ is $-100 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The bond enthalpy of $B_2$ is
1 $100 \mathrm{~kJ} \mathrm{~mol}^{-1}$
2 $50 \mathrm{~kJ} \mathrm{~mol}^{-1}$
3 $200 \mathrm{~kJ} \mathrm{~mol}^{-1}$
4 $150 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Explanation:
Given that, $\mathrm{A}_2: \mathrm{B}_2: \mathrm{AB}=2: 1: 2$ $\Delta \mathrm{H}_{\mathrm{AB}}=-100 \mathrm{kJmol}^{-1}$ The following reaction take place - $\begin{gathered} \mathrm{A}_2+\mathrm{B}_2 \rightarrow 2 \mathrm{AB} \\ \text { or } \quad \frac{1}{2} \mathrm{~A}_2+\frac{1}{2} \mathrm{~B}_2 \rightarrow \mathrm{AB} \end{gathered}$ We can calculate the bond enthalpy of $B_2$ by the give equation - $\begin{array}{ll} & \frac{1}{2} \Delta \mathrm{H}_{\mathrm{A}_2}+\frac{1}{2} \Delta \mathrm{H}_{\mathrm{B}_2}-\frac{1}{2} \Delta \mathrm{H}_{\mathrm{AB}}=-100 \\ \text { or } & \frac{1}{2} \times 2 \mathrm{x}+\frac{\mathrm{x}}{2}-2 \mathrm{x}=-100 \\ \text { or } & \mathrm{x}=200 \mathrm{kJmol}^{-1}\end{array}$
Kerala-CEE-29.08.2021
Thermodynamics
273016
Which one of the following arrangements shows the bonds $\mathrm{H}-\mathrm{H}, \mathrm{C}-\mathrm{C}$ and $\mathrm{Si}-\mathrm{Si}$ in order of increasing bond energy?
$\mathrm{Si}-\mathrm{Si}=(340 \mathrm{~kJ} / \mathrm{mole})$ $\begin{aligned} & \mathrm{C}-\mathrm{C}=(347 \mathrm{~kJ} / \mathrm{mole}) \\ & \mathrm{H}-\mathrm{H}=(432 \mathrm{~kJ} / \mathrm{mole}) \end{aligned}$ Greater the size of the atom, greater is the bond length and less is the bond energy. Si has more size than $\mathrm{C}$ and $\mathrm{H}$. So the bond length between Si-Si is more and the bond dissociation energy will be less. So, the correct order is $\mathrm{Si}-\mathrm{Si}<\mathrm{C}-\mathrm{C}<\mathrm{H}-\mathrm{H}$
SCRA 2012
Thermodynamics
273017
If the mass defect of a nuclide is $3.32 \times 10^{-26} \mathrm{~g}$, its binding energy is MeV.
1 9.31
2 18.62
3 27.93
4 37.24
Explanation:
Given mass defect $=3.32 \times 10^{-26} \mathrm{gm}$. $\text { or } \begin{aligned} \text { mass defect } & =\frac{3.32 \times 10^{-26}}{1.66 \times 10^{-24}} \mathrm{u} \\ & =2 \times 10^{-2} \mathrm{u} \end{aligned}$ $\therefore \quad$ Binding energy $=$ mass defect $\times 931 \mathrm{MeV}$ $=2 \times 10^{-2} \times 931 \mathrm{MeV}$ $=18.62 \mathrm{MeV}$
273036
The energy that opposes dissolution of a solvent is
1 hydration energy
2 lattice energy
3 internal energy
4 bond energy
Explanation:
Lattice energy opposes dissolution of a solvent. Lattice energy is the amount of energy required to dissociate 1 mole of ionic crystals into ions. If hydration energy of solid is more than lattice energy, then solid dissolves.
BITSAT 2006
Thermodynamics
273035
The following is endothermic reaction
1 Decomposition of water
2 Conversion of graphite to diamond
3 Dehydrogenation of ethane to ethylene
4 All of the above
Explanation:
The reaction during which heat is observed known as endothermic reaction. (A) Combustion is always an exothermic procure. (B) Decomposition of $\mathrm{H}_2 \mathrm{O}$ requires energy (C) Dehydrogenation is an endothermic reaction. (D) Graphite is more stable than diamond and energy in needed to convert graphite into diamond.
BCECE-2010
Thermodynamics
273015
Bond enthalpies of $A_2, B_2$ and $A B$ are in the ratio $2: 1: 2$. If bond enthalpy of formation of $A B$ is $-100 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The bond enthalpy of $B_2$ is
1 $100 \mathrm{~kJ} \mathrm{~mol}^{-1}$
2 $50 \mathrm{~kJ} \mathrm{~mol}^{-1}$
3 $200 \mathrm{~kJ} \mathrm{~mol}^{-1}$
4 $150 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Explanation:
Given that, $\mathrm{A}_2: \mathrm{B}_2: \mathrm{AB}=2: 1: 2$ $\Delta \mathrm{H}_{\mathrm{AB}}=-100 \mathrm{kJmol}^{-1}$ The following reaction take place - $\begin{gathered} \mathrm{A}_2+\mathrm{B}_2 \rightarrow 2 \mathrm{AB} \\ \text { or } \quad \frac{1}{2} \mathrm{~A}_2+\frac{1}{2} \mathrm{~B}_2 \rightarrow \mathrm{AB} \end{gathered}$ We can calculate the bond enthalpy of $B_2$ by the give equation - $\begin{array}{ll} & \frac{1}{2} \Delta \mathrm{H}_{\mathrm{A}_2}+\frac{1}{2} \Delta \mathrm{H}_{\mathrm{B}_2}-\frac{1}{2} \Delta \mathrm{H}_{\mathrm{AB}}=-100 \\ \text { or } & \frac{1}{2} \times 2 \mathrm{x}+\frac{\mathrm{x}}{2}-2 \mathrm{x}=-100 \\ \text { or } & \mathrm{x}=200 \mathrm{kJmol}^{-1}\end{array}$
Kerala-CEE-29.08.2021
Thermodynamics
273016
Which one of the following arrangements shows the bonds $\mathrm{H}-\mathrm{H}, \mathrm{C}-\mathrm{C}$ and $\mathrm{Si}-\mathrm{Si}$ in order of increasing bond energy?
$\mathrm{Si}-\mathrm{Si}=(340 \mathrm{~kJ} / \mathrm{mole})$ $\begin{aligned} & \mathrm{C}-\mathrm{C}=(347 \mathrm{~kJ} / \mathrm{mole}) \\ & \mathrm{H}-\mathrm{H}=(432 \mathrm{~kJ} / \mathrm{mole}) \end{aligned}$ Greater the size of the atom, greater is the bond length and less is the bond energy. Si has more size than $\mathrm{C}$ and $\mathrm{H}$. So the bond length between Si-Si is more and the bond dissociation energy will be less. So, the correct order is $\mathrm{Si}-\mathrm{Si}<\mathrm{C}-\mathrm{C}<\mathrm{H}-\mathrm{H}$
SCRA 2012
Thermodynamics
273017
If the mass defect of a nuclide is $3.32 \times 10^{-26} \mathrm{~g}$, its binding energy is MeV.
1 9.31
2 18.62
3 27.93
4 37.24
Explanation:
Given mass defect $=3.32 \times 10^{-26} \mathrm{gm}$. $\text { or } \begin{aligned} \text { mass defect } & =\frac{3.32 \times 10^{-26}}{1.66 \times 10^{-24}} \mathrm{u} \\ & =2 \times 10^{-2} \mathrm{u} \end{aligned}$ $\therefore \quad$ Binding energy $=$ mass defect $\times 931 \mathrm{MeV}$ $=2 \times 10^{-2} \times 931 \mathrm{MeV}$ $=18.62 \mathrm{MeV}$
273036
The energy that opposes dissolution of a solvent is
1 hydration energy
2 lattice energy
3 internal energy
4 bond energy
Explanation:
Lattice energy opposes dissolution of a solvent. Lattice energy is the amount of energy required to dissociate 1 mole of ionic crystals into ions. If hydration energy of solid is more than lattice energy, then solid dissolves.
BITSAT 2006
Thermodynamics
273035
The following is endothermic reaction
1 Decomposition of water
2 Conversion of graphite to diamond
3 Dehydrogenation of ethane to ethylene
4 All of the above
Explanation:
The reaction during which heat is observed known as endothermic reaction. (A) Combustion is always an exothermic procure. (B) Decomposition of $\mathrm{H}_2 \mathrm{O}$ requires energy (C) Dehydrogenation is an endothermic reaction. (D) Graphite is more stable than diamond and energy in needed to convert graphite into diamond.
BCECE-2010
Thermodynamics
273015
Bond enthalpies of $A_2, B_2$ and $A B$ are in the ratio $2: 1: 2$. If bond enthalpy of formation of $A B$ is $-100 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The bond enthalpy of $B_2$ is
1 $100 \mathrm{~kJ} \mathrm{~mol}^{-1}$
2 $50 \mathrm{~kJ} \mathrm{~mol}^{-1}$
3 $200 \mathrm{~kJ} \mathrm{~mol}^{-1}$
4 $150 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Explanation:
Given that, $\mathrm{A}_2: \mathrm{B}_2: \mathrm{AB}=2: 1: 2$ $\Delta \mathrm{H}_{\mathrm{AB}}=-100 \mathrm{kJmol}^{-1}$ The following reaction take place - $\begin{gathered} \mathrm{A}_2+\mathrm{B}_2 \rightarrow 2 \mathrm{AB} \\ \text { or } \quad \frac{1}{2} \mathrm{~A}_2+\frac{1}{2} \mathrm{~B}_2 \rightarrow \mathrm{AB} \end{gathered}$ We can calculate the bond enthalpy of $B_2$ by the give equation - $\begin{array}{ll} & \frac{1}{2} \Delta \mathrm{H}_{\mathrm{A}_2}+\frac{1}{2} \Delta \mathrm{H}_{\mathrm{B}_2}-\frac{1}{2} \Delta \mathrm{H}_{\mathrm{AB}}=-100 \\ \text { or } & \frac{1}{2} \times 2 \mathrm{x}+\frac{\mathrm{x}}{2}-2 \mathrm{x}=-100 \\ \text { or } & \mathrm{x}=200 \mathrm{kJmol}^{-1}\end{array}$
Kerala-CEE-29.08.2021
Thermodynamics
273016
Which one of the following arrangements shows the bonds $\mathrm{H}-\mathrm{H}, \mathrm{C}-\mathrm{C}$ and $\mathrm{Si}-\mathrm{Si}$ in order of increasing bond energy?
$\mathrm{Si}-\mathrm{Si}=(340 \mathrm{~kJ} / \mathrm{mole})$ $\begin{aligned} & \mathrm{C}-\mathrm{C}=(347 \mathrm{~kJ} / \mathrm{mole}) \\ & \mathrm{H}-\mathrm{H}=(432 \mathrm{~kJ} / \mathrm{mole}) \end{aligned}$ Greater the size of the atom, greater is the bond length and less is the bond energy. Si has more size than $\mathrm{C}$ and $\mathrm{H}$. So the bond length between Si-Si is more and the bond dissociation energy will be less. So, the correct order is $\mathrm{Si}-\mathrm{Si}<\mathrm{C}-\mathrm{C}<\mathrm{H}-\mathrm{H}$
SCRA 2012
Thermodynamics
273017
If the mass defect of a nuclide is $3.32 \times 10^{-26} \mathrm{~g}$, its binding energy is MeV.
1 9.31
2 18.62
3 27.93
4 37.24
Explanation:
Given mass defect $=3.32 \times 10^{-26} \mathrm{gm}$. $\text { or } \begin{aligned} \text { mass defect } & =\frac{3.32 \times 10^{-26}}{1.66 \times 10^{-24}} \mathrm{u} \\ & =2 \times 10^{-2} \mathrm{u} \end{aligned}$ $\therefore \quad$ Binding energy $=$ mass defect $\times 931 \mathrm{MeV}$ $=2 \times 10^{-2} \times 931 \mathrm{MeV}$ $=18.62 \mathrm{MeV}$
273036
The energy that opposes dissolution of a solvent is
1 hydration energy
2 lattice energy
3 internal energy
4 bond energy
Explanation:
Lattice energy opposes dissolution of a solvent. Lattice energy is the amount of energy required to dissociate 1 mole of ionic crystals into ions. If hydration energy of solid is more than lattice energy, then solid dissolves.
BITSAT 2006
Thermodynamics
273035
The following is endothermic reaction
1 Decomposition of water
2 Conversion of graphite to diamond
3 Dehydrogenation of ethane to ethylene
4 All of the above
Explanation:
The reaction during which heat is observed known as endothermic reaction. (A) Combustion is always an exothermic procure. (B) Decomposition of $\mathrm{H}_2 \mathrm{O}$ requires energy (C) Dehydrogenation is an endothermic reaction. (D) Graphite is more stable than diamond and energy in needed to convert graphite into diamond.
BCECE-2010
Thermodynamics
273015
Bond enthalpies of $A_2, B_2$ and $A B$ are in the ratio $2: 1: 2$. If bond enthalpy of formation of $A B$ is $-100 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The bond enthalpy of $B_2$ is
1 $100 \mathrm{~kJ} \mathrm{~mol}^{-1}$
2 $50 \mathrm{~kJ} \mathrm{~mol}^{-1}$
3 $200 \mathrm{~kJ} \mathrm{~mol}^{-1}$
4 $150 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Explanation:
Given that, $\mathrm{A}_2: \mathrm{B}_2: \mathrm{AB}=2: 1: 2$ $\Delta \mathrm{H}_{\mathrm{AB}}=-100 \mathrm{kJmol}^{-1}$ The following reaction take place - $\begin{gathered} \mathrm{A}_2+\mathrm{B}_2 \rightarrow 2 \mathrm{AB} \\ \text { or } \quad \frac{1}{2} \mathrm{~A}_2+\frac{1}{2} \mathrm{~B}_2 \rightarrow \mathrm{AB} \end{gathered}$ We can calculate the bond enthalpy of $B_2$ by the give equation - $\begin{array}{ll} & \frac{1}{2} \Delta \mathrm{H}_{\mathrm{A}_2}+\frac{1}{2} \Delta \mathrm{H}_{\mathrm{B}_2}-\frac{1}{2} \Delta \mathrm{H}_{\mathrm{AB}}=-100 \\ \text { or } & \frac{1}{2} \times 2 \mathrm{x}+\frac{\mathrm{x}}{2}-2 \mathrm{x}=-100 \\ \text { or } & \mathrm{x}=200 \mathrm{kJmol}^{-1}\end{array}$
Kerala-CEE-29.08.2021
Thermodynamics
273016
Which one of the following arrangements shows the bonds $\mathrm{H}-\mathrm{H}, \mathrm{C}-\mathrm{C}$ and $\mathrm{Si}-\mathrm{Si}$ in order of increasing bond energy?
$\mathrm{Si}-\mathrm{Si}=(340 \mathrm{~kJ} / \mathrm{mole})$ $\begin{aligned} & \mathrm{C}-\mathrm{C}=(347 \mathrm{~kJ} / \mathrm{mole}) \\ & \mathrm{H}-\mathrm{H}=(432 \mathrm{~kJ} / \mathrm{mole}) \end{aligned}$ Greater the size of the atom, greater is the bond length and less is the bond energy. Si has more size than $\mathrm{C}$ and $\mathrm{H}$. So the bond length between Si-Si is more and the bond dissociation energy will be less. So, the correct order is $\mathrm{Si}-\mathrm{Si}<\mathrm{C}-\mathrm{C}<\mathrm{H}-\mathrm{H}$
SCRA 2012
Thermodynamics
273017
If the mass defect of a nuclide is $3.32 \times 10^{-26} \mathrm{~g}$, its binding energy is MeV.
1 9.31
2 18.62
3 27.93
4 37.24
Explanation:
Given mass defect $=3.32 \times 10^{-26} \mathrm{gm}$. $\text { or } \begin{aligned} \text { mass defect } & =\frac{3.32 \times 10^{-26}}{1.66 \times 10^{-24}} \mathrm{u} \\ & =2 \times 10^{-2} \mathrm{u} \end{aligned}$ $\therefore \quad$ Binding energy $=$ mass defect $\times 931 \mathrm{MeV}$ $=2 \times 10^{-2} \times 931 \mathrm{MeV}$ $=18.62 \mathrm{MeV}$
273036
The energy that opposes dissolution of a solvent is
1 hydration energy
2 lattice energy
3 internal energy
4 bond energy
Explanation:
Lattice energy opposes dissolution of a solvent. Lattice energy is the amount of energy required to dissociate 1 mole of ionic crystals into ions. If hydration energy of solid is more than lattice energy, then solid dissolves.
BITSAT 2006
Thermodynamics
273035
The following is endothermic reaction
1 Decomposition of water
2 Conversion of graphite to diamond
3 Dehydrogenation of ethane to ethylene
4 All of the above
Explanation:
The reaction during which heat is observed known as endothermic reaction. (A) Combustion is always an exothermic procure. (B) Decomposition of $\mathrm{H}_2 \mathrm{O}$ requires energy (C) Dehydrogenation is an endothermic reaction. (D) Graphite is more stable than diamond and energy in needed to convert graphite into diamond.
BCECE-2010
Thermodynamics
273015
Bond enthalpies of $A_2, B_2$ and $A B$ are in the ratio $2: 1: 2$. If bond enthalpy of formation of $A B$ is $-100 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The bond enthalpy of $B_2$ is
1 $100 \mathrm{~kJ} \mathrm{~mol}^{-1}$
2 $50 \mathrm{~kJ} \mathrm{~mol}^{-1}$
3 $200 \mathrm{~kJ} \mathrm{~mol}^{-1}$
4 $150 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Explanation:
Given that, $\mathrm{A}_2: \mathrm{B}_2: \mathrm{AB}=2: 1: 2$ $\Delta \mathrm{H}_{\mathrm{AB}}=-100 \mathrm{kJmol}^{-1}$ The following reaction take place - $\begin{gathered} \mathrm{A}_2+\mathrm{B}_2 \rightarrow 2 \mathrm{AB} \\ \text { or } \quad \frac{1}{2} \mathrm{~A}_2+\frac{1}{2} \mathrm{~B}_2 \rightarrow \mathrm{AB} \end{gathered}$ We can calculate the bond enthalpy of $B_2$ by the give equation - $\begin{array}{ll} & \frac{1}{2} \Delta \mathrm{H}_{\mathrm{A}_2}+\frac{1}{2} \Delta \mathrm{H}_{\mathrm{B}_2}-\frac{1}{2} \Delta \mathrm{H}_{\mathrm{AB}}=-100 \\ \text { or } & \frac{1}{2} \times 2 \mathrm{x}+\frac{\mathrm{x}}{2}-2 \mathrm{x}=-100 \\ \text { or } & \mathrm{x}=200 \mathrm{kJmol}^{-1}\end{array}$
Kerala-CEE-29.08.2021
Thermodynamics
273016
Which one of the following arrangements shows the bonds $\mathrm{H}-\mathrm{H}, \mathrm{C}-\mathrm{C}$ and $\mathrm{Si}-\mathrm{Si}$ in order of increasing bond energy?
$\mathrm{Si}-\mathrm{Si}=(340 \mathrm{~kJ} / \mathrm{mole})$ $\begin{aligned} & \mathrm{C}-\mathrm{C}=(347 \mathrm{~kJ} / \mathrm{mole}) \\ & \mathrm{H}-\mathrm{H}=(432 \mathrm{~kJ} / \mathrm{mole}) \end{aligned}$ Greater the size of the atom, greater is the bond length and less is the bond energy. Si has more size than $\mathrm{C}$ and $\mathrm{H}$. So the bond length between Si-Si is more and the bond dissociation energy will be less. So, the correct order is $\mathrm{Si}-\mathrm{Si}<\mathrm{C}-\mathrm{C}<\mathrm{H}-\mathrm{H}$
SCRA 2012
Thermodynamics
273017
If the mass defect of a nuclide is $3.32 \times 10^{-26} \mathrm{~g}$, its binding energy is MeV.
1 9.31
2 18.62
3 27.93
4 37.24
Explanation:
Given mass defect $=3.32 \times 10^{-26} \mathrm{gm}$. $\text { or } \begin{aligned} \text { mass defect } & =\frac{3.32 \times 10^{-26}}{1.66 \times 10^{-24}} \mathrm{u} \\ & =2 \times 10^{-2} \mathrm{u} \end{aligned}$ $\therefore \quad$ Binding energy $=$ mass defect $\times 931 \mathrm{MeV}$ $=2 \times 10^{-2} \times 931 \mathrm{MeV}$ $=18.62 \mathrm{MeV}$
