Hydrogen cyanide is a highly poisonous, volatile liquid. It can be prepared by the following reaction.
CH4(g) + NH3(g) → HCN(g) + 3 H2(g)
What is the heat of reaction at constant pressure? Use the following thermochemical equations.
N2(g) + 3 H2(g) → 2 NH3(g) ΔH = −91.8 kJ
C(graphite) + 2 H2(g) → CH4(g) ΔH = −74.9 kJ
H2(g) + 2 C(graphite) + N2(g) → 2 HCN(g) ΔH = 270.3 kJ
delta H= 45.9+74.9+135.15=255.95
NH3(g)->N2(g)+3/2H2(G)=+45.9KJ(*1/2)and reversed
CH4(g)->C(graphite)+2H2(g)=+74.9KJ reversed
1/2H2(g)+C(graphite)+1/2N2(g)-> HCN(g)=+135.1KJ
Cancel out C(graphite)and 1/2N2(g) on opposite sides of the reaction.
NH3(g)->H2(g)+1/2H2(g)=+45.9KJ (simplified H2(g)with a factor of 3/2 to 1+1/2=3/2)
CH4(g)-> 2H2(s)=+74.9KJ
1/2H2(g)->HCN(g)=+135.1KJ
cancel out 1/2H2(g) and sum up the reations to obtain.
NH3(g)+CH4(g)->HCN+3H2(g)=+225.9KJ
To find the heat of reaction at constant pressure for the given reaction, we need to use Hess's Law. Hess's Law states that if a reaction can be expressed as the sum of multiple reactions, then the enthalpy change of the overall reaction is the sum of the enthalpy changes of the individual reactions.
First, let's identify the relevant reactions and their enthalpy changes:
1. N2(g) + 3 H2(g) → 2 NH3(g) ΔH = -91.8 kJ (Given)
2. C(graphite) + 2 H2(g) → CH4(g) ΔH = -74.9 kJ (Given)
3. H2(g) + 2 C(graphite) + N2(g) → 2 HCN(g) ΔH = 270.3 kJ (Given)
4. CH4(g) + NH3(g) → HCN(g) + 3 H2(g) (Target Reaction)
We can manipulate the given reactions to match the target reaction:
Multiply reaction (1) by 2:
2 [N2(g) + 3 H2(g) → 2 NH3(g)] → N2(g) + 6 H2(g) → 4 NH3(g) ΔH = -183.6 kJ
Multiply reaction (2) by 2:
2 [C(graphite) + 2 H2(g) → CH4(g)] → 2 C(graphite) + 4 H2(g) → 2 CH4(g) ΔH = -149.8 kJ
Now, we can add these manipulated reactions to obtain the target reaction:
N2(g) + 6 H2(g) + 2 C(graphite) + 4 H2(g) → 4 NH3(g) + 2 CH4(g)
Simplifying the equation:
N2(g) + 10 H2(g) + 2 C(graphite) → 4 NH3(g) + 2 CH4(g)
We can see that this equation matches the stoichiometry of the target reaction. Now, we add the enthalpy changes:
ΔH(target) = ΔH(1) + ΔH(2) + ΔH(3)
= -183.6 kJ + (-149.8 kJ) + 270.3 kJ
= -62.9 kJ
Therefore, the heat of reaction at constant pressure for the given reaction is -62.9 kJ.