Answer:
-625 kcal/mol
Explanation:
The method to solve this question is based on Hess´s law of constant heat of summation which allows us to combine the enthalpies of individual reactions for which we know their enthalpy to obtain the enthalpy change for a desired reaction.
We are asked to calculate the standard enthalpy of formation of combustion of an unbranched alkane :
CnHn+2 unbranched + O₂ ⇒ CO₂ + H₂O ΔcHº = ?
where CnH2n+2 is the general formula for alkanes.
and we are given information for
n C+ (2n + n)/2 H₂ ⇒ CnHn+2 unbranched ΔfHº = -35 kcal/mol (1)
n C+ (2n + n)/2 H₂ ⇒ CnHn+2 branched ΔfHº = -28 kcal/mol (2)
CnHn+2 branched + O₂ ⇒ CO₂ + H₂O ΔcHº = -632 kcal/mol (3)
If we reverse (1) and add it to the sum (2) and (3) we get the desired equation for the combustion of the unbranched alkane:
CnHn+2 unbranched + O₂ ⇒ CO₂ + H₂O
Thus
ΔcHº unbranched = + 35 kcal/mol + (-28 kcal/mol) + (-632 kcal/mol)
= -625 kcal/mol
Answer:
Chemical Changes in Burning Candle: When you light the candle, the wax present near the wick will melt. Wick absorbs the liquid wax. The liquid wax vaporizes due the heat produced by the flame. This wax vapor near to flame burns and gives new substances like Carbon Dioxide, Carbon soot, water vapours, heat and light.
Explanation:
No, don't try, it will explode close to 187 kPa
