Calculate the value of DHo for the reaction 2 CH4 (g) -> C2H6 (g) + H2 (g) given the following thermochemical equations:
C2H2 (g) + H2 (g) -> C2H4 (g) delta = – 175.1 kJ
C2H6 (g) -> C2H4 (g) + H2 (g) Delta = + 136.4 kJ
C2H2 (g) + 3 H2 (g) -> 2 CH4 (g) Delta = – 376.8 kJ
a) + 65.3 kJ
b) + 338.1 kJ
c) – 415.5 kJ
d) + 688.3 kJ
dont understand help!
This is an application of Hess' Law.
You want to arrange the three equations given to obtain the desired equation.
Reverse eqn 3 and 2 and add i eqn 1. When you reverse an equation, change the sign of delta H.
If you follow my directions, see if you get the equation you want and add the delta H values (with changed signs). You should obtain one of the answers.
65.3?
yes
To calculate the value of ΔH° for the given reaction, we need to use Hess's Law, which states that the enthalpy change of a reaction is equal to the sum of the enthalpy changes of the reactions that make up the reaction.
Here's how you can solve this problem step by step:
Step 1: Write down the given reaction and the three thermochemical equations:
Given reaction:
2 CH4 (g) -> C2H6 (g) + H2 (g)
Thermochemical equations:
C2H2 (g) + H2 (g) -> C2H4 (g) ΔH1 = -175.1 kJ
C2H6 (g) -> C2H4 (g) + H2 (g) ΔH2 = +136.4 kJ
C2H2 (g) + 3 H2 (g) -> 2 CH4 (g) ΔH3 = -376.8 kJ
Step 2: Manipulate the equations to match the given reaction:
We need to manipulate the equations so that when we sum them up, the products and reactants match the given reaction.
Equation 1: C2H2 (g) + H2 (g) -> C2H4 (g)
Multiply this equation by 2 so that the number of ethene (C2H4) molecules matches the reactant:
2 C2H2 (g) + 2 H2 (g) -> 2 C2H4 (g) ΔH1' = 2 × ΔH1 = -350.2 kJ
Equation 2: C2H6 (g) -> C2H4 (g) + H2 (g)
No manipulation is needed as it already matches the given reaction:
C2H6 (g) -> C2H4 (g) + H2 (g) ΔH2 = +136.4 kJ
Equation 3: C2H2 (g) + 3 H2 (g) -> 2 CH4 (g)
Multiply this equation by 2 so that the number of methane (CH4) molecules matches the products:
2 C2H2 (g) + 6 H2 (g) -> 4 CH4 (g) ΔH3' = 2 × ΔH3 = -753.6 kJ
Step 3: Sum up the manipulated equations:
Adding the manipulated equations will cancel out the intermediates and leave us with the desired reaction:
-350.2 kJ + 136.4 kJ - 753.6 kJ = -967.4 kJ
Step 4: Determine the value of ΔH° for the given reaction:
Since we want ΔH° for the given reaction, we need to flip the sign of the sum calculated in the previous step:
ΔH° = - (-967.4 kJ) = +967.4 kJ
Therefore, the value of ΔH° for the given reaction is +967.4 kJ.
Note that none of the provided answer choices matches the calculated value exactly.