Ethylene glycol, HOCH2CH2OH, is used as antifreeze. It is produced from ethylene oxide, C2H4O, by the following reaction.
C2H4O(g) + H2O(l) → HOCH2CH2OH(l)
Use Hess's law to obtain the enthalpy change for this reaction from the following enthalpy changes.
2 C2H4O(g) + 5 O2(g) → 4 CO2(g) + 4 H2O(l) ΔH = −2612.2 kJ
HOCH2CH2OH(l) + 5/2 O2(g) → 2 CO2(g) + 3 H2O(l) ΔH = −1189.8 kJ
Well, let's clown around with Hess's law!
First, notice that the first equation is doubled, so let's double that enthalpy change:
2 C2H4O(g) + 10 O2(g) → 8 CO2(g) + 8 H2O(l) ΔH = −5224.4 kJ
Next, we need to reverse the second equation, so the sign of the enthalpy change will change:
−(HOCH2CH2OH(l) + 5/2 O2(g) → 2 CO2(g) + 3 H2O(l)) ΔH = 1189.8 kJ
Now we add these two equations together:
2 C2H4O(g) + 10 O2(g) → 8 CO2(g) + 8 H2O(l) ΔH = −5224.4 kJ
−(HOCH2CH2OH(l) + 5/2 O2(g) → 2 CO2(g) + 3 H2O(l)) ΔH = 1189.8 kJ
2 C2H4O(g) + 10 O2(g) − HOCH2CH2OH(l) − 5/2 O2(g) → 8 CO2(g) + 8 H2O(l) − 2 CO2(g) − 3 H2O(l)
ΔHtotal = −5224.4 kJ + 1189.8 kJ
Simplifying the equation:
2 C2H4O(g) + 5 O2(g) + HOCH2CH2OH(l) → 6 CO2(g) + 5 H2O(l)
ΔHtotal = -4034.6 kJ
So, the enthalpy change for the reaction is approximately -4034.6 kJ. Now that's one hot and cold reaction!
To obtain the enthalpy change for the reaction C2H4O(g) + H2O(l) → HOCH2CH2OH(l) using Hess's law, we need to manipulate and combine the given enthalpy changes to cancel out the desired reactants and products.
Let's analyze the desired reaction and the given reactions:
Desired reaction: C2H4O(g) + H2O(l) → HOCH2CH2OH(l)
Given reactions:
1. 2 C2H4O(g) + 5 O2(g) → 4 CO2(g) + 4 H2O(l) ΔH = −2612.2 kJ
2. HOCH2CH2OH(l) + 5/2 O2(g) → 2 CO2(g) + 3 H2O(l) ΔH = −1189.8 kJ
To use Hess's law, we need to multiply and/or manipulate the given reactions to align the stoichiometry with the desired reaction. Here's how we go about it:
1. Reverse the second reaction:
-2 CO2(g) - 3 H2O(l) → HOCH2CH2OH(l) + 5/2 O2(g) ΔH = 1189.8 kJ
2. Multiply the first reaction by 2 to match the number of moles of C2H4O in the desired reaction and flip the equation:
-4 CO2(g) - 4 H2O(l) → 2 C2H4O(g) + 5 O2(g) ΔH = 5224.4 kJ
Now, let's sum up the modified reactions to get the desired reaction:
-4 CO2(g) - 4 H2O(l) + -2 CO2(g) - 3 H2O(l) → 2 C2H4O(g) + 5 O2(g) + HOCH2CH2OH(l) + 5/2 O2(g)
Simplifying the equation:
-6 CO2(g) - 7 H2O(l) → 2 C2H4O(g) + 15/2 O2(g) + HOCH2CH2OH(l)
Finally, add up the enthalpy changes of the modified reactions to obtain the enthalpy change for the desired reaction:
ΔH = (5224.4 kJ) + (1189.8 kJ) = 6414.2 kJ
Therefore, the enthalpy change for the reaction C2H4O(g) + H2O(l) → HOCH2CH2OH(l) using Hess's law is 6414.2 kJ.