Question

From the data below, calculate the total heat (in j) needed to convert 0.782 mol of gaseous ethanol at 300.0°c and 1 atm to liquid ethanol at 25.0°c and 1 atm

Answer 1

Answer:

You must remove $\text{50.6 kJ}$ .

Explanation:

There are three heat transfers in this process:

Total heat = cool the vapour + condense the vapour + cool the liquid  

       q          =           q₁            +                q₂                   +           q₃

       q          =       nC₁ΔT₁        +          nΔHcond             +        nC₂ΔT₂

Let's calculate these heat transfers separately.

Data:

You don't give "the data below", so I will use my best estimates from the NIST Chemistry WebBook. You can later substitute your own values.

C₁ = specific heat capacity of vapour = 90 J·K⁻¹mol⁻¹

C₂ = specific heat capacity of liquid   = 115 J·K⁻¹mol⁻¹

ΔHcond = -38.56 kJ·mol⁻¹

Tmax = 300   °C

  b.p. =   78.4 °C

Tmin =   25.0 °C

n = 0.782 mol

Calculations:

ΔT₁ = 78.4 - 300 = -221.6 K

q₁ = 0.782 × 90 × (-221.6) = -15 600 J = -15.60 kJ

q₂ = 0.782 × (-38.56) = -30.15 kJ

ΔT = 25.0 - 78.4 = -53.4 K

q₃ = 0.782 × 115 × (-53.4) = -4802 J = 4.802 kJ

q = -15.60 - 53.4 - 4.802 = -50.6 kJ

You must remove $\text{50.6 kJ}$ of heat to convert the vapour to a gas.