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A vapor volume of 1.17 L forms when a sample of liquid acetonitrile, CH3CN, absorbs 1.00 kJ of heat at its normal boiling point (81.6 °C and 1 atm). What is Hvap in kilojoules per mole of CH3CN?​

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pstnonsonjoku

The heat of vaporization of CH3CN is obtained from the question as 25 KJ/mol.

We can obtain the number of moles from;

PV = nRT

P = 1 atm

V = 1.17 L

n = ?

R = 0.082 atm LK-1mol-1

T = 81.6 °C + 273 = 354.6 K

n = PV/RT

Substituting values;

n =  1 atm × 1.17 L/0.082 atm LK-1mol-1 × 354.6 K

n = 0.04 moles

Using; q = n·ΔHv

q = Heat absorbed

n = number of moles

Hv  = Heat of vaporization

ΔHv = q/n

ΔHv = 1.00 × 10^3/0.04

ΔHv = 25 KJ/mol

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