Question

Determine the heat of evaporation of carbon disulfide, CS 2() - CS 2(g) given the enthalpies of reaction below. C(s) + 2 S(s)-CS() C(s) + 2 S(s) - CS2(g) ArHº = +89.4 kJ/mol-rxn ArHº = -116.7 kJ/mol-rxn

Answer 1

ANSWER:

Question 1

CS2 (1) CS2 (9)

The evaporation of carbon disulfide could be expressed as the sum of two reactions:

CS2 (1) + C(s) +25(S)

C(s) +2S(s) → CS2 (9)

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CS2 (1) CS2 (9)

Then, the heat of evaporation is expressed as

Hvap = A# Rr1 + A# Rr2

• Reaction 1 is the inverse reaction of

C(s) +2s(s) + CS2 (1) AHR = +89.4kJ/mol

then, reaction 1 has a ΔH = -89.4 kJ/mol

Finally, the heat of evaporation of carbon disulfide is

Hvap = A# Rr1 + A# Rr2

A Hvap = (-89.4 kJ/mol) +(+116.7 kJ/mol)

AHvap = +27.3 kJ/mol

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Question 2

Only the compounds in its standard state has an enthalpy of formation equal to zero. An element in standard state means that is the natural state of element (at 25º C, 1 atm), par example:

• the standard state of molecular oxygen (O2) is gas
• the standard state of mercury (Hg) is liquid
• the standard state of carbon (C) is solid
• Then, C(s), O2 (g) and Hg has a standard enthalphy of formation equals to zero

In standard conditions the bromine is found as Br2 and it is a liquid. Then, the Br2 (l) has zero standard enthalphy of formation at 25 ºC.