Which of the following statements is correct for the spontaneous adsorption of a gas?

1.	∆ S is negative and therefore,  ∆ H should be highly positive
2.	∆ S is negative and therefore,  ∆ H should be highly negative
3.	∆ S is positive and therefore,  ∆ H should be negative
4.	-∆ S is positive and therefore,  ∆ H should also be highly positive

Given the following reaction:
\(4H(g)\)→  \(2 H_{2}\)\((g)\)
The enthalpy change for the reaction is -869.6 kJ. The dissociation energy of the H-H bond is:
1. -869.6 kJ
2. +434.8kJ
3. +217.4kJ
4. -434.8 kJ

From the following bond energies:
H—H bond energy: 431.37 kJ mol^-1 
C=C bond energy: 606.10 kJ mol^-1 
C—C bond energy: 336.49 kJ mol^-1 
C—H bond energy: 410.50 kJ mol^-1 
Enthalpy for the reaction, 

will be:

The values of ΔH and ΔS for the given reaction are 170 kJ and 170 JK^-1, respectively.
C_(graphite) + CO₂(g)→2CO(g) 
This reaction will be spontaneous at:

1. 710 K
2. 910 K
3. 1110 K
4. 510 K

The bond energy of H—H and Cl-Cl is 430 kJ mol^-1 and 240 kJ mol^-1 respectively
and ΔH_f for HCl is -90 kJ mol^-1. The bond enthalpy of HCl is:

1. 290 $kJ$ $mo{l}^{-1}$

2. 380 $kJ$ $mo{l}^{-1}$

3. 425 $kJ$ $mo{l}^{-1}$

4. 245 $kJ$ $mo{l}^{-1}$

Two moles of an ideal gas is heated at a constant pressure of one atmosphere from 27$°$C to 127$°$C. If ${\mathrm{C}}_{\mathrm{v}, \mathrm{m}}=20+{10}^{-2} \mathrm{T} {\mathrm{JK}}^{-1} {\mathrm{mol}}^{-1},$ then q and $△\mathrm{U}$ for the process are respectively:

1.  6362.8 J, 4700 J

2.  3037.2 J, 4700 J

3.  7062.8 J, 5400 J

4.  3181.4 J, 2350 J

Calculate ${△}_{\mathrm{f}}\mathrm{H}°$ (in kJ/mol) for ${\mathrm{Cr}}_2{\mathrm{O}}_3$ from the ${△}_{\mathrm{r}}\mathrm{G}°$ and the $\mathrm{S}°$ values provided at 27$°$C

$4\mathrm{Cr}\left(\mathrm{s}\right)+3{\mathrm{O}}_2\left(\mathrm{g}\right)$ $\to$ $2{\mathrm{Cr}}_2{\mathrm{O}}_3\left(\mathrm{s}\right);$ $$
${△}_{\mathrm{r}}\mathrm{G}°=-2093.4$ $\mathrm{kJ}/\mathrm{mol}$

$\mathrm{S}°\left(\mathrm{J}/\mathrm{K}\right)$ $\mathrm{mol}$ $:$ $\mathrm{S}°\left(\mathrm{Cr},\right)$ $\mathrm{s}=24;$ $$
$\mathrm{S}°\left({\mathrm{O}}_2,\right)$ $\mathrm{g}=205;$ $$ $$ $\mathrm{S}°\left({\mathrm{Cr}}_2{\mathrm{O}}_3,\right)$ $\mathrm{s}=81$

1.  -2258.1 kJ/mol

2.  -1129.05 kJ/mol

3.  -964.35 kJ/mol

4.  None of the above

The correct statement for a reversible process in a state of equilibrium is:
1. $∆$G = - 2.30RT log K
2. $∆$G = 2.30RT log K
3. $∆$G^o = - 2.30RT log K
4. $∆$G^o = 2.30RT log K