Calculate delta H for the reaction below using Hess's Law:

C2H6 ---> C2H2 + 2H2
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C2H2 + 5/2O2 --> 2CO2 + H2O (H = -1300 kJ)
2H2 + O2 --> 2H2O (H= -572 kJ)
2C2H6 + 7O2 --> 4CO2 + 6H2O (H= -3120 kJ)

Multiply eqn 1 by 2 and reverse. Multiply the delta H value by 2 and change the sign.

Multiply eqn 2 by 2 and reverse it. Multiply the delta H value by 2 and change the sign.

Add the new eqn 1 + eqn 2 along with eqn 3 and you get twice the equn you want. Add new delta H values for total and take half of it.

842.5

To calculate ΔH for the reaction C2H6 → C2H2 + 2H2 using Hess's Law, we need to use the given equations and manipulate them to obtain the desired reaction.

First, we need to reverse the direction of the second equation:

2CO2 + H2O → C2H2 + 5/2O2

Then, we multiply the first equation by 2 and the third equation by 2:

2(C2H2 + 5/2O2 → 2CO2 + H2O)
2(C2H6 → C2H2 + 2H2)

Now, we can add the equations together, canceling out any species that appear on both sides of the reaction:

2C2H6 + 7O2 → 4CO2 + 6H2O
2C2H2 + 5O2 → 4CO2 + 2H2O

Next, we want to eliminate the extra O2 and H2O in the second equation. We can do this by doubling the first equation and canceling out the appropriate species:

4C2H2 + 10O2 → 8CO2 + 4H2O
2C2H2 + 5O2 → 4CO2 + 2H2O

Finally, we subtract the second equation from the first equation:

2C2H6 + 7O2 - (2C2H2 + 5O2) → 4CO2 + 6H2O - (4CO2 + 2H2O)

Simplifying the equation, we get:

2C2H6 + 7O2 - 2C2H2 - 5O2 → 4CO2 + 6H2O - 4CO2 - 2H2O

Cancelling out the like terms:

2C2H6 - 2C2H2 + 7O2 - 5O2 → 6H2O - 2H2O

Simplifying further:

2C2H6 - 2C2H2 + 2O2 → 4H2O

The ΔH for the reaction C2H6 → C2H2 + 2H2 can be calculated by summing the enthalpies from the reactions we used to manipulate the equations:

ΔH = -(3120 kJ) - (-1300 kJ) - (-572 kJ)
ΔH = -3120 kJ + 1300 kJ - 572 kJ
ΔH = -2392 kJ

Therefore, the ΔH for the reaction C2H6 → C2H2 + 2H2 is -2392 kJ.

To calculate delta H for the reaction using Hess's Law, we need to find a series of reactions whose enthalpy values can be combined to get the desired reaction.

Given reactions:
1. C2H2 + 5/2O2 --> 2CO2 + H2O (H = -1300 kJ)
2. 2H2 + O2 --> 2H2O (H= -572 kJ)
3. 2C2H6 + 7O2 --> 4CO2 + 6H2O (H= -3120 kJ)

We need to rearrange and combine equations 1 and 2 in order to cancel out the compounds present in the desired reaction.

Step 1: Reverse equation 1 because we want to produce C2H2 instead of consuming it:
2CO2 + H2O --> C2H2 + 5/2O2 (H = +1300 kJ)

Step 2: Multiply equation 2 by 2 to balance the number of H2O, so that we can cancel it out with the H2O in equation 1:
4H2 + 2O2 --> 4H2O (H = -1144 kJ)

Step 3: Multiply equation 1 by 2, so that we have 4H2O to cancel out with the 4H2O in step 2:
4CO2 + 2H2O --> 2C2H2 + 5O2 (H = -2600 kJ)

Step 4: Now, we can add equation 3 with the rearranged and combined equations from steps 2 and 3:
2C2H6 + 7O2 --> 4CO2 + 6H2O (H = -3120 kJ)
+4CO2 + 2H2O --> 2C2H2 + 5O2 (H = -2600 kJ)
+4H2 + 2O2 --> 4H2O (H = -1144 kJ)

Combine these equations to get the desired reaction:
2C2H6 + 7O2 --> 2C2H2 + 5O2 + 4CO2 + 6H2O

Step 5: Add up the enthalpy values of the combined equations to get the delta H for the desired reaction:
-3120 kJ + (-2600 kJ) + (-1144 kJ) = -6864 kJ

Therefore, the delta H for the reaction C2H6 ---> C2H2 + 2H2 is -6864 kJ.