An oxyacetylene torch is a tool that mixes and burns oxygen and acetylene to produce an extremely hot flame. This tool is used to cut steel or weld iron and other metals. The temperature of the film can reach 3480 oC. The burning of acetylene is given by the thermochemical equation:
C2H2 (g) + 3/2 CO2 (g) —> CO2 (g) + H2O (l) ΔH = - 1301.1 kJ
For the reaction,
2CO2 (g) + 2H2O (l) —> 2 C2H2 (g) + 3 CO2 (g)
what is the ΔH for the reaction?
Well, if we add up the given equations, we can cancel out the CO2 and H2O on both sides to find the overall reaction:
C2H2 (g) + 3/2 CO2 (g) —> CO2 (g) + H2O (l) ΔH = -1301.1 kJ
2 CO2 (g) + 2 H2O (l) —> 2 C2H2 (g) + 3 CO2 (g)
By canceling out the CO2 and H2O, we get:
2 C2H2 (g) —> 3 CO2 (g)
So, the ΔH for the reaction would still be -1301.1 kJ. And I must say, that's quite an impressive amount of heat for a reaction! Just imagine if we could harness that energy to cook our meals or warm up our homes. But alas, we'll have to stick with our regular stoves and heaters for now.
To find the ΔH for the given reaction, we need to use the given thermochemical equation and apply the concept of Hess's Law. Hess's Law states that if a reaction can be expressed as a series of steps, the overall enthalpy change (ΔH) for the reaction is the sum of the enthalpy changes for each step.
Let's break down the reaction into two steps:
Step 1: Multiply the given equation by 2 to get the same number of CO2 molecules on both sides:
2 C2H2 (g) + 3 CO2 (g) -> 4CO2 (g) + 2H2O (l)
Step 2: Reverse the given equation:
- 2CO2 (g) - 2H2O (l) -> 2 C2H2 (g) + 3 CO2 (g)
Now, we can compare the overall reaction with the combination of these two steps:
2 C2H2 (g) + 3 CO2 (g) -> 4CO2 (g) + 2H2O (l)
- 2CO2 (g) - 2H2O (l) -> 2 C2H2 (g) + 3 CO2 (g)
By canceling out the common terms on both sides, we can write the overall reaction:
2 C2H2 (g) -> 4CO2 (g) + 2H2O (l)
To find the ΔH for the overall reaction, we need to sum up the enthalpy changes from each step.
Step 1: ΔH1 = -1301.1 kJ (from the given thermochemical equation)
Step 2: Since we reversed the given equation, the sign of ΔH should be reversed. So ΔH2 = -(-1301.1 kJ) = +1301.1 kJ.
Add the two enthalpy changes together:
ΔH = ΔH1 + ΔH2 = -1301.1 kJ + 1301.1 kJ = 0 kJ
Therefore, the ΔH for the reaction "2CO2 (g) + 2H2O (l) -> 2 C2H2 (g) + 3 CO2 (g)" is 0 kJ.
To find the ΔH for the given reaction, we can use the thermochemical equation provided.
The equation C2H2 (g) + 3/2 CO2 (g) → CO2 (g) + H2O (l) has a ΔH of -1301.1 kJ, which means that 1301.1 kJ of energy is released during this reaction.
Now, let's analyze the reaction we are interested in:
2CO2 (g) + 2H2O (l) → 2C2H2 (g) + 3CO2 (g)
We can see that this reaction is the reverse of the given thermochemical equation, so we need to reverse the sign of ΔH.
Therefore, the ΔH for the reaction
2CO2 (g) + 2H2O (l) → 2C2H2 (g) + 3CO2 (g)
is +1301.1 kJ.