Question

Calculate the total heat (in J) needed to covert 0.333 mol of gaseous ethanol (CH₃CH₂OH) at 300 °C and 1 atm to liquid ethanol at 25.0°C and 1 atm. Use the following data: Normal bp = 78.5 °C, ΔH_vap = 40.5 kJ mol^-1, c_gas = 1.43 J g^-1 °C^-1, c_liq = 2.45 J g^-1 °C^-1.

Answer 1

Hi, thank you for your question. Here we have two things to consider. First, gaseous ethanol is converted to liquid ethanol at the boiling point and second, liquid ethanol is cooled from its boiling point to 25.0 C.

As condensation is exothermic, heat will be releases in the process, please note that.

mass of ethanol is 0.333 x 46 = 15.3 g.

heat from condensation at 78.5 C is:

, we are using negative signs since the process is exothermic.

= {-15.3 x 1.43 x (300-78.5)} - 40.5 = -4886.7 J

Now the heat released when liquid ethanol is cooled from 78.5 C to 25.0 C

= -15.3 x 2.45 x (78.5-25.0) = -2005.4 J

total heat is the sum of both, so (-4886.7) + (-2005.4) = -6892.1 J.

you can also write that J of heat will be releases during the process.

If you need any other information, please do not hesitate to ask.