Thermodynamics ...

"The change of enthalpy of a chemical reaction is the same whether the reaction takes place in one step or in several steps". This law was presented by:

Consider the Born-Haber cycle for the formation of an ionic compound given below and identify the compound $$\left(Z\right)$$ formed.

If enthalpies of formation for $$C_2H_4(g)$$, $$CO_2(g)$$ and $$H_2O(l)$$ at $$25^o$$C and $$1$$ atm pressure be $$52$$, $$-394$$ and $$-286$$ $$kJ \ mol^{-1}$$ respectively, then the enthalpy of combustion of $$C_2H_4(g)$$ will be:

Spontaneous reactions are :

 For the reactions,

A mixture of two moles of carbon monoxide and one mole of oxygen, in a closed vessel is ignited to convert the carbon monoxide to carbon dioxide. If $$\Delta$$H is the enthalpy change and $$\Delta$$E is the change in internal energy, then:

The enthalpies of combustion of $${C}_{(graphite)}$$ and $${C}_{(diamond)}$$ are $$-393.5$$ and $$-395.4kJ/mol$$ respectively. The enthalpy of conversion of $${C}_{(graphite)}$$ to $${C}_{(diamond)}$$ in $$kJ/mol$$ is:

If $$\Delta {H}_{f}^{o}$$ for $${H}_{2}{O}_{2}(l)$$ and $${H}_{2}O(l)$$ are $$-188kJ$$ $${mol}^{-1}$$ and $$-286kJ$$ $${mol}^{-1}$$, what will be the enthalpy change of the reaction?
$$2{H}_{2}{O}_{2}(l)\rightarrow 2{H}_{2}O(l)+{O}_{2}(g)$$

From the thermochemical reactions,
$${C}_{(graphite)}+\cfrac{1}{2}{O}_{2}\rightarrow CO;\Delta H=-110.5kJ$$
$$CO+\cfrac{1}{2}{O}_{2}\rightarrow {CO}_{2};\Delta H=-283.2kJ$$
the heat of reaction of $${C}_{(graphite)}+{O}_{2}\rightarrow {CO}_{2}$$ is:

Given:
$$C(s)+{O}_{2}(g)\rightarrow {CO}_{2}(g)+94.2kcal$$
$${H}_{2}(g)+\cfrac{1}{2}{O}_{2}(g)\rightarrow {H}_{2}O(l)+68.3kcal$$
$${CH}_{4}+2{O}_{2}(g)\rightarrow {CO}_{2}(g)+2{H}_{2}O(l)+210.8kcal$$
The heat of formation of methane in $$kcal$$ will be: