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

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

Give 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 following 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 ${\mathrm{C}}_{\mathrm{v},<mspace\; width="1em"></mspace>\mathrm{m}}=20+{10}^{-2}<mspace\; width="1em"></mspace>\mathrm{T}<mspace\; width="1em"></mspace>{\mathrm{JK}}^{-1}<mspace\; width="1em"></mspace>{\mathrm{mol}}^{-1},<mspace\; width="1em"></mspace>$ 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

\(\begin{array}{ll} 4 C r(s)+3 O_2(g) \rightarrow 2 C r_2 O_3(s), \\ \Delta_r G^{\circ}=-2093.4 k J / m o l \\ S^{\circ}(\mathrm{J} / / \mathrm{K} \mathrm{~mol}): S^{\circ}(C r, s)=24, \\ S^{\circ}\left(O_2, g\right)=205, \quad S^{\circ}\left(C r_2 O_3, s\right)=81 \end{array}\)

1.  -2258.1 kJ/mol
2.  -1129.05 kJ/mol
3.  -964.35 kJ/mol
4.  None of the above

The standard heat of combustion of propane is –2220.1 kJ mol^–1. The standard heat of vaporisation of liquid water is 44.0 kJ mol^–1. The enthalpy change for the reaction is–

C₃H₈ (g) + 5O₂ (g) $\to$ 3CO₂ (g) + 4H₂O(g)