Thermodynamics ...

Use the following data to calculate second electron ainity of oxygen, i.e., for the process
$$O^{-}(g) + e^{-}(g)  \rightarrow O^{2-}(g)$$
Is the $$O^{2-}$$ ion stable in the gas phase?.Why is it stable in solid MgO?
Heat of sublimation of $$Mg(s) = + 147.7 kJ mol^{-1}$$
Ionisation energy of Mg(g) to form
$$Mg^{2+}(g) = + 2189.0 kJ mol^{-1}$$
Bond dissociation energy for $$O_2 = + 498.4 kJmol^{-1}$$
First electron affinity of $$O(g) = - 141.0 kJ mol^{-1}$$
Heat formation of $$MgO(s) = -601.7 kJ mol^{-1}$$
Lattice energy of $$MgO = -3791.0 kJ mol^{-1}$$

The energy change for the alternating reaction that yields chlorine sodium $$(Cl^{+}Na^{-})$$ will be:

$$2Na(s)\, +\, Cl_2(g)\,\rightarrow\, 2Cl^{+}Na^{-}(s)$$

Given that:

Lattice energy of $$NaCl\,=\,-787\, kJ\,mol^{-1}$$

Electron affinity of $$Na\,=\,-52.9\, kJ\, mol^{-1}$$

Ionisation energy of $$Cl\, =\, +\,1251\, kJ\, mol^{-1}$$

BE of $$Cl_2\,=\,244\, kJ\, mol^{-1}$$

Heat of sublimation of $$Na(s)\, =\,107.3\, kJ\, mol^{-1}$$

$$\Delta H_f(NaCl)\, =\,-411\, kJ\, mol^{-1}$$.

Determine $${ \Delta  }{ U }^{ o }$$ at $$300K$$ for the following reaction using the listed enthalpies of reaction:

Using the listed information calculate $${ \Delta  }_{ r }{ G }^{ o }$$ (in kJ/mol) at $${ 27 }^{ o }C$$.
$${ Co }_{ 3 }{ O }_{ 4 }(s)+4CO(g)\longrightarrow 3Co(s)+4{ CO }_{ 2 }(g)$$

Given : At $$300K$$ , $$\Delta { { H }^{ o } }_{ f }(kJ/mol) $$ are $$-891,\ -110.5,\ 0.0$$ and $$-393.5$$ respectively.
           $${ S }^{ o }(J/K.mol)$$ are $$102.5,\ 197.7, \ 30.0$$ and $$213.7$$ respectively.

The lattice energy of NaCl(s) using the following data will be:

heat of sublimation of $$Na(s)\,=\,S$$

$$(IE)_1$$ of $$Na\,(g)\,=\,I$$

bond dissociation energy of $$Cl_2\,(g)\,=\,D$$

electron affinity of $$Cl\,(g)\,=\,-E$$

Fixed mass of an ideal gas contained in a $$24.63L$$ sealed rigid vessel at $$1$$ atm is heated from $$-{ 73 }^{ o }C$$ to $${ 27 }^{ o }C$$. Calculate change in Gibb's energy if entropy of gas is a function of temperature as $$S=2+{ 10 }^{ -2 }T(J/K)$$: (Use $$1$$ atm $$L=0.1kJ$$)

If the Gibbs free energy change when 1 mole of $$\displaystyle NaCl$$ is dissolved in water at 298K is $$x$$ kJ, then -$$1000x$$ is
Given that,
(a) Lattice energy of $$\displaystyle NaCl$$ = $$\displaystyle 778\ kJ{ mol }^{ -1 }$$
(b) Hydration energy of $$\displaystyle NaCl = -774.3\ kJ{ mol }^{ -1 }$$
(c) Entropy change at $$\displaystyle 298K=43\ J{ mol }^{ -1 }$$.

 For the reaction $$2Al + Fe_2O_3\longrightarrow 2Fe+Al_2O_3$$, the standard heat enthalpy of $$Fe_2O_3$$ and $$Al_2O_3$$ are $$-196.5$$ and $$ -399.1$$ kcal respectively. $$\Delta H^{\circ}$$ for the reaction is:

The reaction that proceeds in the forward direction is:

From the following data of $$\Delta H$$, of the following reactions: