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
To obtain the enthalpy change for the reaction C2H4O(g) + H2O(l) → HOCH2CH2OH(l) using Hess's law, we need to manipulate the given reactions so that they sum up to the desired reaction. Here are the steps:
1. Given reactions:
a. 2 C2H4O(g) + 5 O2(g) → 4 CO2(g) + 4 H2O(l) ΔH = -2612.2 kJ
b. HOCH2CH2OH(l) + 5/2 O2(g) → 2 CO2(g) + 3 H2O(l) ΔH = -1189.8 kJ
2. Reverse Reaction b:
Multiply the reaction b by -1 to reverse the direction:
-1 × (HOCH2CH2OH(l) + 5/2 O2(g) → 2 CO2(g) + 3 H2O(l)) ΔH = 1189.8 kJ
3. Multiply Reaction a by 2:
Multiply the reaction a by 2 to obtain the same number of moles of C2H4O as in the desired reaction:
2 × (2 C2H4O(g) + 5 O2(g) → 4 CO2(g) + 4 H2O(l)) ΔH = -2 × 2612.2 kJ = -5224.4 kJ
4. Multiply Reaction b by 4:
Multiply the reversed reaction b by 4 to obtain the same number of moles of HOCH2CH2OH as in the desired reaction:
4 × (-1) × (HOCH2CH2OH(l) + 5/2 O2(g) → 2 CO2(g) + 3 H2O(l)) ΔH = 4 × 1189.8 kJ = 4759.2 kJ
5. Manipulate the reactions to obtain the desired reaction:
Add the manipulated reactions (step 3 and step 4) to cancel out the intermediates and obtain the desired reaction:
-5224.4 kJ + 4759.2 kJ = -464.2 kJ
Therefore, the enthalpy change for the reaction C2H4O(g) + H2O(l) → HOCH2CH2OH(l) is -464.2 kJ.
To obtain the enthalpy change for the reaction of ethylene oxide with water to produce ethylene glycol, we can use Hess's law. Hess's law states that the overall enthalpy change of a reaction is independent of the pathway taken, and is equal to the sum of the enthalpy changes of the individual steps of the reaction.
First, we need to write the balanced equation for the reaction we want to find the enthalpy change for:
C2H4O(g) + H2O(l) → HOCH2CH2OH(l)
Now, let's break down the reaction into two steps using the given enthalpy changes:
Step 1: 2 C2H4O(g) + 5 O2(g) → 4 CO2(g) + 4 H2O(l) ΔH = -2612.2 kJ
This reaction represents the combustion of ethylene oxide to form carbon dioxide and water.
Step 2: HOCH2CH2OH(l) + 5/2 O2(g) → 2 CO2(g) + 3 H2O(l) ΔH = -1189.8 kJ
This reaction represents the combustion of ethylene glycol to form carbon dioxide and water.
To obtain the desired reaction, we need to reverse Step 2 and multiply it by a factor to cancel out the number of moles of ethylene glycol. Since there are two moles of ethylene glycol in the desired reaction and three moles in Step 2, we will multiply Step 2 by 2/3.
Now, let's add the two steps together:
2 C2H4O(g) + 5 O2(g) → 4 CO2(g) + 4 H2O(l) ΔH = -2612.2 kJ
(2/3)(HOCH2CH2OH(l) + 5/2 O2(g) → 2 CO2(g) + 3 H2O(l)) ΔH = (2/3)(-1189.8 kJ)
Combining these equations, we cancel out the reactants and products that appear in both steps:
C2H4O(g) + H2O(l) + (2/3)(HOCH2CH2OH(l) + 5/2 O2(g) → 2 CO2(g) + 3 H2O(l) ΔH = -2612.2 kJ + (2/3)(-1189.8 kJ)
Simplifying the equation:
C2H4O(g) + H2O(l) → HOCH2CH2OH(l) ΔH = -2612.2 kJ + (-791.2 kJ)
Finally, we calculate the enthalpy change for the reaction:
ΔH = -2612.2 kJ + (-791.2 kJ) = -3403.4 kJ
Therefore, the enthalpy change for the reaction of ethylene oxide with water to produce ethylene glycol is -3403.4 kJ.