Calculate the standard enthalpy of formation (\(\Delta_fH^\circ\)) for 2 moles of liquid benzene (\(\mathrm{{C_6H_6}_{(l)}}\)) at 25°C, based on the given thermodynamic data.
Given Data:
\(\Delta _c\mathrm H\mathrm{(C_6H_6}_\mathrm{(l)})~= -3264.6~ \mathrm{~kJ} / \mathrm{mol}\)
\(\Delta _c\mathrm H\mathrm{(C}_\mathrm{(s)})~= -393.5~ \mathrm{~kJ} / \mathrm{mol}\)
\(\Delta _f\mathrm H\mathrm{(H_2O}_\mathrm{(l)})~= -285.83~ \mathrm{~kJ} / \mathrm{mol}\)

1. \(-~92.22 \mathrm{~kJ} / \mathrm{mol}\)
2. \(-46.11 \mathrm{~kJ} / \mathrm{mol}\)
3. \(+~92.22 \mathrm{~kJ} / \mathrm{mol}\)
4. \(+46.11 \mathrm{~kJ} / \mathrm{mol}\)
Subtopic:  Thermochemistry |
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Given :

C(Graphite)+O2(g)CO2(g);
rH°=-393.5KJmol-1;
H2(g)+12O2(g)H2O(I);
rH°=-285.8KJmol-1;
CO2(g)+2H2O(I)CH4(g)+2O2(g);
rH°=+890.3KJmol-1     

Based on the above thermochemical equations, the value of rH° at 298 K for the reaction

C(Graphite)+2H2(g)CH4(g)willbe:

1. 85.2 kJ/mol

2. 15.2 kJ/mol

3. -74.8 kJ/mol

4. 74.8 kJ/mol

Subtopic:  Thermochemistry |
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The enthalpy of combustion of four allotropic forms of element 'X' are given as:
Allotropic forms \(\Delta_\text{comb}H^\circ\)(kJ/mol)
A. -270.3
B. -189.1
C. -390.5
D. -465.0
The most stable allotropic form of element 'X' is:

1.  A
2.  B
3.  C
4.  D
Subtopic:  Thermochemistry |
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