At 25°C the following heats of reaction are known:
2C2H2 + 5O2 ---> 4CO2 + 2H2O; ΔH = -2600 kJ
C + O2 ---> CO2 ; ΔH = -394 kJ
2H2 + O2 ---> 2H2O; ΔH = -572 kJ
At the same temperature, calculate ΔH for the following reaction:
2C + H2 ---> C2H2; ΔH = ??
I don't understand how to solve this, what equation would i use?
Well, you've stumbled upon a heat of reaction party! It's all about the enthalpy change, baby! To solve this particular problem, you can use a process called Hess's Law. No, it has nothing to do with the Hess Truck that's released every holiday season.
Hess's Law states that if a reaction can be expressed as a combination of other reactions, then the overall enthalpy change is equal to the sum of the individual enthalpy changes.
In this case, you can make use of the reactions you already have their enthalpy changes known. Let's break the desired reaction down into smaller steps and hope they don't fall apart like a clown's attempt at juggling:
Step 1: 2C2H2 + 5O2 ---> 4CO2 + 2H2O; ΔH = -2600 kJ
Step 2: C + O2 ---> CO2; ΔH = -394 kJ
Step 3: 2H2 + O2 ---> 2H2O; ΔH = -572 kJ
Now, to get our desired reaction: 2C + H2 ---> C2H2
We can reverse Step 2 by multiplying it by -1. We do this because we want the carbon dioxide to become carbon, and we want to form acetylene (C2H2) instead.
Step 2 (reversed): CO2 ---> C + O2; ΔH = 394 kJ
Now, all we have to do is add up these steps to get the overall reaction! Here it is:
Step 1 + Step 2 (reversed) + Step 3 = 2C + H2 ---> C2H2
Now let's add up the enthalpy changes:
-2600 kJ + 394 kJ + (-572 kJ) = ΔH
Don't be afraid of some simple math here! Give it a shot and calculate ΔH for this reaction.
To solve the problem, you can use Hess's Law. Hess's Law states that if a reaction can be expressed as a sum of two or more other reactions, then the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the individual reactions.
To apply Hess's Law, you need to manipulate the given reactions in order to obtain the desired reaction. Here's how you can do it step-by-step:
1. Reverse the second reaction:
CO2 ---> C + O2 ; ΔH = +394 kJ
2. Double the first reaction and react it with the reversed second reaction:
2(C2H2) + 10(O2) ---> 8(CO2) + 4(H2O) + 2(C) + 10(O2) ; ΔH = -2(-2600 kJ) + 2(+394 kJ)
Simplifying: 2(C2H2) + 10(O2) ---> 8(CO2) + 4(H2O) + 2(C) ; ΔH = 5200 kJ + 788 kJ
ΔH = 5988 kJ
3. Cancel out the common compounds on both sides of the equation:
2(C2H2) + 10(O2) ---> 8(CO2) + 4(H2O) + 2(C)
4. Divide the equation by 2 to obtain the desired reaction:
C2H2 + 5(O2) ---> 4(CO2) + 2(H2O) + C ; ΔH = 2994 kJ
Therefore, ΔH for the reaction 2C + H2 ---> C2H2 is 2994 kJ.
To solve for the ΔH of the given reaction, you can use the concept of Hess's Law. Hess's Law states that the enthalpy change (ΔH) of a reaction is independent of the pathway taken to obtain the final products, as long as the initial and final conditions are the same.
In this case, we need to determine the ΔH for the reaction: 2C + H2 ---> C2H2
To find the ΔH for this reaction, we can use the known ΔH values for the given reactions and apply them to the reaction we want to solve.
First, let's rearrange and reverse the reaction we want to solve so that it matches one of the known reactions:
C2H2 ---> 2C + H2
Now, we can apply the known ΔH values to this rearranged reaction. We will use the known reactions in such a way that the overall reaction cancels out the compounds that are not present in the rearranged reaction:
2C2H2 + 5O2 ---> 4CO2 + 2H2O; ΔH = -2600 kJ
(1/2)(2C2H2 + 5O2) ---> (1/2)(4CO2 + 2H2O); ΔH = (1/2)(-2600 kJ) = -1300 kJ
C + O2 ---> CO2 ; ΔH = -394 kJ
2C + 2O2 ---> 2CO2; ΔH = 2(-394 kJ) = -788 kJ
2H2 + O2 ---> 2H2O; ΔH = -572 kJ
2H2 + O2 ---> 2H2O; ΔH = -572 kJ
Now, we can add the equations together:
(1/2)(-2600 kJ) + (-788 kJ) + (-572 kJ) ---> ΔH
-1300 kJ - 788 kJ - 572 kJ ---> ΔH
ΔH = -2660 kJ
Therefore, the ΔH for the reaction 2C + H2 ---> C2H2 is -2660 kJ.
you need to combine the given reactions
... some partial and/or reversed
... to get the desired reaction
2CO2 + H2O --> C2H2 + 5/2 02 ΔH = 1300 kJ
2C + 2O2 --> 2CO2 ΔH = -788 kJ
H2 + 1/2 O2 --> H20 ΔH = -286 kJ
add the 3 reactions, along with the ΔH's
... cancel like terms on opposite sides of the result