How to draw hess' cycle for this question ?
carbon, hydrogen and ethane each burn exothermically in an excess of air. c(s) + o2(g) → co2(g) ahⓡ =-393.7 kj mol: 1. h2(g) + % o2(g) → h2o(1) ah® --285.9 kj mol. c2h4(g) + 302(g) → 2co2(g) + 2h2o(1) ah® --1411.0 kj mol? . use the data to calculate the standard enthalpy change of formation, ah® in kj mol? , of ethene at 298 k 2c(s) + 2h2(g) → c2h4(8)

Answer : The standard enthalpy of formation of ethylene is, 51.8 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of will be,

The intermediate balanced chemical reaction will be,

(1)    

(2)    

(3)    

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equations, we get :

(1)    

(2)    

(3)    

The expression for enthalpy of formation of will be,

Therefore, the standard enthalpy of formation of ethylene is, 51.8 kJ/mole

air

in a tire you fill a tire with air

then a couple days later you check the air pressure in the tire

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