Question

Ethylene oxide (EO) is prepared by the vapor-phase oxidation of ethylene. Its main uses are in the preparation of the antifreeze ethylene glycol and in the production of polyethylene terephthalate), which is used to make beverage bottles and fibers. Pure EO vapor can decompose explosively.
Liquid EO has ΔHf= -77.4 kJ/mOl and ΔHvap= 569.4 J/g.

a. Calculate ΔHrxn, for the gas-phase reaction.
b. Due to external heating, the vapor decomposes at 10 bar and 930°C in a distillation column. What is the final temperature if the average specific heat capacity of the products is 2.5 J/g°C?

Answer 1

Answer:

a. Δ$H^0_{rxn} = -108.0\frac{kJ}{mol}$

b. 320.76° C

Explanation:

a.)

we can solve this type of question (i.e calculate Δ$H^0_{rxn}$ , for the gas-phase reaction )  using the Hess's Law.

Δ$H^0_{rxn}$ =  $E_{product} deltaH^0_{t}-E_{reactant} deltaH^0_{t}$

Given from the question, the table below shows the corresponding  Δ$H^0_{t}(kJ/mol)$ for each compound.

Compound                    $H^0_{t}(kJ/mol)$

Liquid EO                       -77.4

$CH_4_(g_)$                            -74.9                

$CO_(g_)$                              -110.5

If we incorporate our data into the above previous equation; we have:

Δ$H^0_{rxn}$ = (-110.5 kJ/mol + (-74.9 kJ/mol) ) - (-77.4 kJ/mol)

          =   $-108.0 \frac{kJ}{mol}$

b.)

We are to find the final temperature if the average specific heat capacity of the products is 2.5 J/g°C

Given that:

the specific heat capacity (c) = 2.5 J/g°C

$T_{initial}$ = 93.0°C   &

the  enthalpy of vaporization  (Δ$H^0_{vap}$) = 569.4 J/g

If, we recall; we will remember that; Specific Heat Capacity is the amount of heat needed to raise the temperature of one gram of a substance by one kelvin.

∴ the specific heat capacity (c) is given as =  $\frac{Heat(q)}{mass*changeintemperature(T_{initial}-T_{final})}$

Let's not forget as well, that  Δ$H^0_{vap}$ = $\frac{q}{mass}$

If we substitute  Δ$H^0_{vap}$  for  $\frac{q}{mass}$ in the above equation, we have;

specific heat capacity (c) = $\frac{deltaH^0_{vap}}{T_{final}-T_{initial}}$

Making ($T_{final}- T_{initial}$) the subject of the formula; we have:

$T_{final}- T_{initial}$  = $\frac{delat H^0_{vap}}{specificheat capacity}$

$(T_{final}-93.0^0C)=\frac{569.4J/g}{2.5J/g^0C}$

$T_{final}=\frac{569.4J/g}{2.5J/g^0C}+93.0^0C$

         = 227.76°C +93.0°C

          = 320.76°C

∴ we can thereby conclude that the final temperature = 320.76°C