Asked Nov 1, 2019

Use standard enthalpies of formation to determine ΔHorxn for:

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

Enter in kJ.


Expert Answer

Step 1

Given information:


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2NH, (g)30, (g) +2CH (g)2HCN (g) 6H,0(g)

Step 2

The chemical equation for enthalpy of formation of product and reactants are represented as follows:



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5H, (g)+C(s, gr) HCNΝ(s).. ... . . . . Ηf (1) = 135.1 k/mol ... ..6 Ε H,(8), (8)HO(1). 0,(8)HO( .ΔΗf (2 ) - -285.83 k/mol N, (8)+5 Η, (s) > NH, (ε).... . .ΔΗ'f (3 ) = -46.11 kJmol ο, (s) -0 (g)........ C(s, gr)+2H, (g)CH, (8) ........ . .ΔΗ=0 .ΔΗf (4 ) - -74.84 kJmol

Step 3

Applying Hess Law and plugging the values of enthalpy of formation of product...


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ΔΗΣΔΗof (products ) -ΣΔΗf (reac tants ) ΔΗ[2 ΔΗf(1) + 6 ΔΗΙ (2)]-[2ΔΗ (3) +2xΔΗf (4)] [2x135.1+6x-285.83] kJ/mol-[2x-46.11+2x -74.81] kJ/mol =-1202.94 kJ/mol


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