Calculate the change of enthalpy for the reaction CH4 (g) + NH3 (g) --> HCN (g) +3H2 (g) from the following reactions:

Reaction 1: N2 (g) + 3H2 (g) --> 2NH3 (g) Change in enthalpy: -91.8 kJ/mol

Reaction 2: C (s, graphite) + 2H2 (g) --> CH4 (g) Change in enthalply: -74.9 kJ/mol

Reaction 3: H2 (g) + 2C (s, graphite) + N2 (g) --> 2HCN (g) Change in enthalpy: +270.3 kJ/mol

Include the following:

The numerical answer with correct units.
State which reactions, if any, you had to "Flip".
State which reactions you had to multiply, if any, to get the correct amount of the compound.

So this is a Hess' Law problem, the CH₄ (g) + NH₃ (g) --> HCN (g) + 3H₂ (g) is what we want the other reactions to reflect. I usually set up problems like these like this in order to determine which reaction needs a coefficient change:

N₂ + 3H₂ --> 2NH₃ (ΔH=-91.8)

C + 2H₂ --> CH₄ (ΔH=-74.9

H₂ + 2C + N₂ --> 2HCN (ΔH=270.3)

CH₄ + NH₃ --> HCN + 3H₂

(I left out the states because it'll make the math easier) So, we want things to cancel out, meaning some of the reactants and products need to change places in order to do so. For the first reaction, we'd want to multiply the coefficients by [tex]\frac{1}{2}[/tex] in order to have it cancel out with the other reactions. For the third reaction, we'd want to we'd want to switch the products/reactants and multiply the coefficients by [tex]\frac{1}{2}[/tex] . Keep in mind whatever we do to the equation, we do to the ΔH. Should look like:

[tex]\frac{1}{2}[/tex]N₂ + [tex]\frac{3}{2}[/tex]H₂ --> NH₃ (ΔH=-45.9)

C + 2H₂ --> CH₄ (ΔH=-74.9)

HCN --> [tex]\frac{1}{2}[/tex]H₂ + C + [tex]\frac{1}{2}[/tex]N₂ (ΔH=-135)

CH₄ + NH₃ --> HCN + 3H₂

Everything cancels, so that means we can add all the ΔH, which should be -255.8 kj/mol, but we also change the sign in order to reflect what's happening in the reaction. (Sorry this is so long)

Abdulazeez10 Abdulazeez10 · Answer 2 · 2021-11-04T18:18:33+01:00

The enthalpy for the reaction, ΔH rxn is 255.95 kJ/mol

From the question,

We are to calculate the change in enthalpy for the reaction

CH₄(g) + NH₃(g) → HCN(g) +3H₂(g)

From the given reactions

Reaction 1: N₂(g) + 3H₂(g) → 2NH₃(g) Change in enthalpy: -91.8 kJ/mol

Reaction 2: C(s, graphite) + 2H₂(g) → CH₄(g) Change in enthalply: -74.9 kJ/mol

Reaction 3: H₂(g) +2C(s, graphite) +N₂(g) → 2HCN (g) Change in enthalpy: +270.3 kJ/mol

First, flip reactions 1 and 2 to get reaction 4 and 5 respectively

Reaction 4: 2NH₃(g) → N₂(g) + 3H₂(g) ΔHo : 91.8 kJ/mol

Reaction 5: CH₄(g) → C(s, graphite) + 2H₂(g) ΔHo : 74.9 kJ/mol

Now, multiply reactions 4 and 3 by half (1/2) to get 6 and 7 respectively

Reaction 6: NH₃(g) → ¹/₂N₂(g) + ³/₂H₂(g) ΔHo : 45.9 kJ/mol

Reaction 7: ¹/₂H₂(g) +C(s, graphite) +¹/₂N₂(g) → HCN (g) ΔHo : +135.15 kJ/mol

Now,

Add reactions 5, 6, and 7 together

Reaction 5: CH₄(g) → C(s, graphite) + 2H₂(g) ΔHo : 74.9 kJ/mol

Reaction 6: NH₃(g) → ¹/₂N₂(g) + ³/₂H₂(g) ΔHo : 45.9 kJ/mol

Reaction 7: ¹/₂H₂(g) +C(s, graphite) +¹/₂N₂(g) → HCN(g) ΔHo : +135.15 kJ/mol

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CH₄(g) + NH₃(g) → HCN(g) + 3H₂(g) ΔH rxn = 255.95 kJ/mol

Hence, the enthalpy for the reaction, ΔH rxn is 255.95 kJ/mol

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