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.

255.8 kj/mol

Explanation:

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 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 . Keep in mind whatever we do to the equation, we do to the ΔH. Should look like:

N₂ + H₂ --> NH₃             (ΔH=-45.9)

C + 2H₂ --> CH₄                (ΔH=-74.9)

HCN --> H₂ + C + 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)

b) mutualism

hope this : )

b. examine and describe cells