Thermodynamics ...

A reaction occurs spontaneously if:

The heat of combustion for $$C$$, $$H_{2}$$ and $$CH_{4}$$ are $$-349$$ kJ/mol, $$-241.8$$ kJ/mol and $$-906.7$$ kJ/mol respectively. The heat of formation in kJ/mol of $$CH_{4}$$ is :

The change in free energy accompanied by the isothermal reversible expansion of 1 mole of an ideal gas when it doubles its volume is $$\Delta G_{1}$$. The change in free energy accompanied by a sudden isothermal irreversible doubling volume of 1 mole of the same gas is $$\Delta G_{2}$$. The ratio of $$\Delta G_{1}$$ and $$\Delta G_{2}$$ is :

Given that
$$Zn + \frac{1}{2} O_2  \rightarrow ZnO\quad\Delta{H}= + 84000  cal$$      ........... 1
$$Hg+\frac{1}{2} O_2  \rightarrow HgO\quad \Delta{H} =+ 21700  cal$$       ...........  2

Consider the following data: $$\Delta_fH^o(N_2H_4, l)=50kJ/mol, \Delta_fH^o(NH_3, g)=-46 kJ/mol$$,
$$B.E. (N-H)=393 kJ/mol$$ and $$B.E. (H-H)=436 kJ/mol$$,
$$\Delta_{vap}H(N_2H_4, l)=18 kJ/mol$$
Calculate the $$N-N$$ bond energy in kJ/mol for $$N_2H_4$$.

The values of $$\Delta H$$ and $$\Delta S$$ for the reaction, C(graphite) + $$CO_2(g)\rightarrow 2CO(g)$$ are 170 kJ and $$170\; JK^{-1}$$ respectively. This reaction will be spontaneous at:

A 1000 gm sample of water is reacted with an equimolar amount of CaO (both at same initial temperature of $$25^{\circ}C$$). Assuming the container to be adiabatic, the final temperature of the product is approx :

If $$\Delta H{_{f}}^{0}$$ for $$Ag^{+} (\infty$$  diluted), $$NO{_{3}}^{-} (\infty$$ diluted), $$Cl^{-} (\infty$$ diluted) and $$AgCl_{(S)}$$ are $$-105.579, -207.36, -167.159$$ and $$-127.068$$ respectively. Calculate the enthalpy change for the reaction $$AgNO_{3(aq)}+HCl_{(aq)}\rightarrow AgCl_{(S)}+HNO_{3} (aq)$$.

Following are the thermochemical reactions:
$$C$$ (graphite) $$+ \dfrac{1}{2}O_2  \rightarrow CO \ ;  \Delta H = - 110.5$$ kJ/mol
$$CO + \dfrac{1}{2} O_2  \rightarrow CO_2 \ ;  \Delta H = - 283.2$$ kJ/mol

Find the heat change in the reaction
$$NH_3(g) + HCl (g)   \rightarrow NH_4 Cl(s)$$ from the following data
$$NH_3 (g) + aq  \rightarrow NH_3 (aq),$$                    $$\Delta H = - 8.4  K.  Cal.$$
$$HCl(g) + aq  \rightarrow HCl(aq),$$                     $$\Delta H = -17.3 K.  Cal.$$
$$NH_3(aq) + HCl(aq) \rightarrow NH_4 Cl(aq),  \Delta H = - 12.5  K. Cals.$$
$$NH_4 Cl(s) + aq \rightarrow NH_4 Cl(aq),                \Delta H = + 3.9  K.  Cal.$$