Answer:
a. Δ![H^0_{rxn} = -108.0\frac{kJ}{mol}](https://tex.z-dn.net/?f=H%5E0_%7Brxn%7D%20%3D%20-108.0%5Cfrac%7BkJ%7D%7Bmol%7D)
b. 320.76° C
Explanation:
a.)
we can solve this type of question (i.e calculate Δ
, for the gas-phase reaction ) using the Hess's Law.
Δ
= ![E_{product} deltaH^0_{t}-E_{reactant} deltaH^0_{t}](https://tex.z-dn.net/?f=E_%7Bproduct%7D%20deltaH%5E0_%7Bt%7D-E_%7Breactant%7D%20deltaH%5E0_%7Bt%7D)
Given from the question, the table below shows the corresponding Δ
for each compound.
Compound ![H^0_{t}(kJ/mol)](https://tex.z-dn.net/?f=H%5E0_%7Bt%7D%28kJ%2Fmol%29)
Liquid EO -77.4
-74.9
-110.5
If we incorporate our data into the above previous equation; we have:
Δ
= (-110.5 kJ/mol + (-74.9 kJ/mol) ) - (-77.4 kJ/mol)
= ![-108.0 \frac{kJ}{mol}](https://tex.z-dn.net/?f=-108.0%20%5Cfrac%7BkJ%7D%7Bmol%7D)
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
= 93.0°C &
the enthalpy of vaporization (Δ
) = 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})}](https://tex.z-dn.net/?f=%5Cfrac%7BHeat%28q%29%7D%7Bmass%2Achangeintemperature%28T_%7Binitial%7D-T_%7Bfinal%7D%29%7D)
Let's not forget as well, that Δ
= ![\frac{q}{mass}](https://tex.z-dn.net/?f=%5Cfrac%7Bq%7D%7Bmass%7D)
If we substitute Δ
for
in the above equation, we have;
specific heat capacity (c) = ![\frac{deltaH^0_{vap}}{T_{final}-T_{initial}}](https://tex.z-dn.net/?f=%5Cfrac%7BdeltaH%5E0_%7Bvap%7D%7D%7BT_%7Bfinal%7D-T_%7Binitial%7D%7D)
Making (
) the subject of the formula; we have:
= ![\frac{delat H^0_{vap}}{specificheat capacity}](https://tex.z-dn.net/?f=%5Cfrac%7Bdelat%20H%5E0_%7Bvap%7D%7D%7Bspecificheat%20capacity%7D)
![(T_{final}-93.0^0C)=\frac{569.4J/g}{2.5J/g^0C}](https://tex.z-dn.net/?f=%28T_%7Bfinal%7D-93.0%5E0C%29%3D%5Cfrac%7B569.4J%2Fg%7D%7B2.5J%2Fg%5E0C%7D)
![T_{final}=\frac{569.4J/g}{2.5J/g^0C}+93.0^0C](https://tex.z-dn.net/?f=T_%7Bfinal%7D%3D%5Cfrac%7B569.4J%2Fg%7D%7B2.5J%2Fg%5E0C%7D%2B93.0%5E0C)
= 227.76°C +93.0°C
= 320.76°C
∴ we can thereby conclude that the final temperature = 320.76°C