Diethyl ether, (CH3CH2)2O, used as a solvent for extraction of organic compounds from aqueous solutions, has a high vapor pressure, which makes it a potential fire hazard in laboratories in which it is used. How much energy is released when 100.0 g is cooled from 53.0°C to 10.0°C? Normal boiling point 34.5°C Heat of vaporization 351 J/g Specific heat of (CH3CH2)2O(l) 3.74 J/g • °C Specific heat of (CH3CH2)2O(g) 2.35 J/g • °C

4.86 x 10⁴ J are released

Explanation:

We need to calculate the heat released  when diethyl ether at 53 ºC is condensed and cooled to a temperature of 10 ªC.

So the process is divided in three steps:

1.- Cool the gas to the normal boiling point

2.-Phase change from gas to liquid

3.- Cool the liquid from 34.5 ºC to 10 ºC

ΔH for step 1:

ΔH₁ = m x csp x ΔT

where m is the mass, csp is the specific heat and ΔT is the change in temperature ( Tfinal - Tinicial ). Here we will be using the specific heat for (CH3CH2)2O(g), 2.35 J/g·ºC.

= 100 g x 2.35 J/g·ºC x ( 34.5 -53  )ºC =  -4.35 x 10³ J

ΔH for step 2:

ΔH₂ condensation  = mC heat of vaporization

= 100.0 g x 351 J/g = -3.51 x 10⁴ J

ΔH₃ for step 3:

ΔH₃ = m x csp x ΔT

Here we will the specific heat for the liquid since now we have the ether in the liquid phase.

ΔH =  100 g x 3.74 J/g·ºC x ( 10.0 - 34.5 )ºC = - 9.16 x 10³ J

The total energy change is:

ΔHtotal = ΔH₁ + ΔH₂ + ΔH₃ =   -4.35 x 10³ J + ( -3.51 x 10⁴ J )= +  (- 9.16 x 10³ J)

ΔHtotal =-4.86 x 10⁴ J

The sign is negative since heat is released.

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