For a particular isomer of c 8 h 18 , the combustion reaction produces 5113.3 kj of heat per mole of c 8 h 18 ( g ) consumed, under standard conditions. c 8 h 18 ( g ) + 25 2 o 2 ( g ) ⟶ 8 co 2 ( g ) + 9 h 2 o ( g ) δ h ∘ rxn = − 5113.3 kj / mol what is the standard enthalpy of formation of this isomer of c 8 h 18 ( g ) ?

ΔH°f(C₈H₁₈(g)) = -210.9 kJ/mol

Explanation:

Let's consider the combustion of C₈H₁₈.

C₈H₁₈(g) + 25/2 O₂(g) ⟶ 8 CO₂(g) + 9 H₂O(g) ΔH°rxn = − 5113.3 kJ

We can calculate the standard enthalpy of formation of C₈H₁₈(g) using the following expression.

ΔH°rxn = 8 mol × ΔH°f(CO₂(g)) + 9 mol × ΔH°f(H₂O(g)) - 1 mol × ΔH°f(C₈H₁₈(g)) - 25/2 mol × ΔH°f(O₂(g))

1 mol × ΔH°f(C₈H₁₈(g)) = 8 mol × ΔH°f(CO₂(g)) + 9 mol × ΔH°f(H₂O(g)) - 25/2 mol × ΔH°f(O₂(g)) - ΔH°rxn

1 mol × ΔH°f(C₈H₁₈(g)) = 8 mol × (-393.5 kJ/mol) + 9 mol × (-241.8 kJ/mol) - 25/2 mol × 0 kJ/mol - (− 5113.3 kJ)

ΔH°f(C₈H₁₈(g)) = -210.9 kJ/mol

answer: find it ur self

explanation:

sodium (na), for example, has one electron in its valence shell. this is an unstable state because that valence shell needs eight electrons to be full. in order to fill its valence shell, find a way to add seven electrons to its valence shell