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german
3 years ago
13

The standard heat of formation of a branched alkane is -35 kcal/mol; the standard heat of formation of the unbranched version of

alkane (same molecular formula) is -28 kcal/mol. Finally, the standard enthalpy of combustion of the branched alkane is -632 kcal/mol. Given this information, what is the standard enthalpy of combustion of the unbranched alkane in terms of kcal/mol to the nearest ones?
Chemistry
1 answer:
AURORKA [14]3 years ago
5 0

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

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We're only talking about molar mass and time (t) here so we'll just concentrate on \sqrt{\frac{M2}{M1} } = \frac{t2}{t1}. Notice how the molar mass and time are on the same position, recall effusion is when gas escapes from a container through a small hole. The time it takes it to leave depends on the molar mass. If the gas is heavy, like Xe, it would take a longer time (4.83 minutes). If it was light it would leave in less time, that gives us somewhat an idea what our element could be, we know that it's atleast an element before Xenon.

Let's plug everything in and solve for M2. I chose M2 to be the unknown here because it's easier to have it basically as a whole number already.

\sqrt{\frac{M2}{131} } = \frac{2.29}{4.83}

The square root is easier to deal with if you take it out in the first step, so let's remove it by squaring each side by 2, the opposite of square root essentially.

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