Question

Use standard enthalpies of formation to determine ΔHorxnfor:

2NH3(g) + 3O2(g) + 2CH4(g) → 2HCN(g) + 6H2O(g)

Enter in kJ.

Answer 1

The standard enthalpy of formation is the enthalpy change (at 100 kPa and usually 25 °C (298.15 K)) for the formation of 1 mole of a substance in itsstandard state from its constituent elements in their standard states (usually the most stable form of the element).

These standard enthalpies can be found on the data tables (usually at the back of your book or u can look them up on google). By exploiting theconservation of energy (look up Hess's Law) you can use these data to determine the standard enthalpy of other reactions.

(1) Your reaction is

2NH3(g) + 3O2(g) + 2CH4(g) --> 2HCN(g) + 6H2O(g)) .....enthaply of reaction, ?H°

Work out enthalpies of formation of products &reactants:

Formation Equations:- Products
1/2H2(g) + C(s, gr) + 1/2N2(g) --> HCN.....enthaply of formation ?H°f(1)
H2(g) + 1/2O2(g) --> H2O(l) .....enthaply of formation, ?H°f(2)

Formation Equations: Reactants
1/2N2(g) + 3/2H2(g) --> NH3(g) .....enthaply of formation, ?H°f(3)
[O2(g) --> O2(g) .....enthaply of formation = 0 kJ mol-¹ by defintion]
C(s, gr) + 2H2(g) --> CH4(g) enthaply of formation, ?H°f(4)

from data tables:
?H°f(1) = 135.1 kJ mol-¹
?H°f(2) = -285.83 kJ mol-¹
?H°f(3) = -46.11 kJ mol-¹
?H°f(4) = -74.81 kJ mol-¹

Using Hess's Law, ?H° = S?H°f(products) - S?H°f(reactants) with the correct stoichiometry:
?H° = [2 x ?H°f(1) + 6 x ?H°f(2)] - [2 x ?H°f(3) + 2 x ?H°f(4)]

Substituting gives ?H° = -1203 kJ mol-¹ (to 4 s.f.).

use the same approach for this to give ?H° = -138.2 kJ mol-¹ (to 4 s.f.).

Answer 2

The standard enthalpy of formation is the enthaply change (at 100 kPa and usually 25 °C (298.15 K)) for the formation of 1 mole of a substance in itsstandard state from its constituent elements in their standard states (usually the most stable form of the element).

These standard enthalpies have been determined and are tabulated in data tables. By exploiting the conservation of energy (look up Hess's Law) you canuse these data to determine the standard enthalpy of other reactions.

(1) Your reaction is

2NH3(g) + 3O2(g) + 2CH4(g) --> 2HCN(g) + 6H2O(g)) .....enthaply of reaction, ΔH°

Work out enthalpies of formation of products &reactants:

Formation Equations:- Products
1/2H2(g) + C(s, gr) + 1/2N2(g) --> HCN.....enthaply of formation ΔH°f(1)
H2(g) + 1/2O2(g) --> H2O(l) .....enthaply of formation, ΔH°f(2)

Formation Equations: Reactants
1/2N2(g) + 3/2H2(g) --> NH3(g) .....enthaply of formation, ΔH°f(3)
[O2(g) --> O2(g) .....enthaply of formation = 0 kJ mol-¹ by defintion]
C(s, gr) + 2H2(g) --> CH4(g) enthaply of formation, ΔH°f(4)

where (from data tables):
ΔH°f(1) = 135.1 kJ mol-¹
ΔH°f(2) = -285.83 kJ mol-¹
ΔH°f(3) = -46.11 kJ mol-¹
ΔH°f(4) = -74.81 kJ mol-¹

Apply Hess's Law, ΔH° = ΣΔH°f(products) - ΣΔH°f(reactants) with the correct stoichiometry:
ΔH° = [2 x ΔH°f(1) + 6 x ΔH°f(2)] - [2 x ΔH°f(3) + 2 x ΔH°f(4)]

Substituting gives ΔH° = -1203 kJ mol-¹ (to 4 s.f.).