1)Which thermochemical equation and data does not agree with the other two written?
a) 2NO (g)+ O2 (g) ->2 NO2 (g) deltaH=-169.8
b) NO (g) + 1/2 O2 (g) -> NO2 (g) delta H = -56.6
c) 4 NO2 (g) -> 4 NO (g) + 2 O2 (g) delta H = +226.4
d)all three equations are in thermochemical agreement.
I chose answer D because all seem correct.
2)Calculate the value of delta H for the reaction 2 CH4 (g) -> C2H6 (g) + H2 (g) given the following thermochemical equations:
C2H2 (g) + H2 (g) -> C2H4 delta H=-175.1
C2H6(g) -> C2H4 (g) + H2 (g) delta H=+136.4
C2H2 (g) + 3 H2 (g) -> 2 CH4 (g) delta H= -376.8
The answer choices are +65.3, +338.1, -415.5, +688.3
I get Delta H as +65.3. Below is how I did it. Please let me know if incorrect.
C2H4 + H2 --> C2H6 Delta H = -136.4 (flipped)
2CH4 --> C2H2 + 3H2 Delta H = +376.8 (flipped)
C2H2 + H2 --> C2H4 Delta H = -175.1
--------------------------------------------
2CH4 --> C2H6 + H2 Delta H = +65.3
2 looks ok to me.
1 is not correct. a is the bad one.
Note that c is 4xb but a is not 2xb
For the problem you worked for Hannah, just a note that it is not necessary to convert to kelvin. Tfinal0-Tinitial is the same in degrees C and degrees K since 1 degree C = 1 degree K.
Using kelvin is not wrong; it just takes longer.
1) You are correct, the answer is (d) all three equations are in thermochemical agreement. All three equations are balanced and provide the correct values for the enthalpy change (delta H).
2) Your calculation is correct. By flipping and adding/reversing the equations, you have correctly determined the overall delta H for the reaction 2 CH4 (g) -> C2H6 (g) + H2 (g) to be +65.3. Well done!
1) To determine which thermochemical equation and data does not agree with the other two, we need to compare the reaction stoichiometry and the delta H values.
a) 2NO (g) + O2 (g) -> 2NO2 (g) delta H = -169.8
b) NO (g) + 1/2 O2 (g) -> NO2 (g) delta H = -56.6
c) 4NO2 (g) -> 4NO (g) + 2O2 (g) delta H = +226.4
First, let's check the stoichiometry:
- In equation a), there are 2 moles of NO reacting with 1 mole of O2 to produce 2 moles of NO2.
- In equation b), there is 1 mole of NO reacting with 1/2 mole of O2 to produce 1 mole of NO2.
- In equation c), there are 4 moles of NO2 reacting to produce 4 moles of NO and 2 moles of O2.
Next, let's compare the delta H values:
- The delta H value for equation a) is -169.8.
- The delta H value for equation b) is -56.6.
- The delta H value for equation c) is +226.4.
Since equation c) requires 4 moles of NO2 to produce 4 moles of NO and 2 moles of O2, while equations a) and b) only require 2 moles of NO2 to produce NO and O2, it can be concluded that equation c) does not agree with the other two equations. Therefore, the correct answer is (c).
2) To calculate the value of delta H for the given reaction, we can use the given thermochemical equations and apply the concept of Hess's Law, which states that the heat change in a chemical reaction is independent of the pathway taken.
Given thermochemical equations:
C2H2 (g) + H2 (g) -> C2H4 (g) delta H = -175.1
C2H6 (g) -> C2H4 (g) + H2 (g) delta H = +136.4
C2H2 (g) + 3H2 (g) -> 2 CH4 (g) delta H = -376.8
To find the delta H for the reaction 2 CH4 (g) -> C2H6 (g) + H2 (g), we need to manipulate the given equations to match the desired reaction. Here's how to do it:
1. Flip equation 2: C2H4 (g) + H2 (g) -> C2H6 (g) with delta H = -136.4 (flipped sign).
2. Flip equation 3: 2 CH4 (g) -> C2H2 (g) + 3H2 (g) with delta H = +376.8 (flipped sign).
Now, let's combine the equations:
1. C2H4 (g) + H2 (g) -> C2H6 (g) delta H = -136.4 (flipped)
2. 2 CH4 (g) -> C2H2 (g) + 3 H2 (g) delta H = +376.8 (flipped)
3. C2H2 (g) + H2 (g) -> C2H4 (g) delta H = -175.1
Add equations 1 and 2 together, taking into account the stoichiometric coefficients:
C2H4 (g) + H2 (g) + 2 CH4 (g) -> C2H6 (g) + 3 H2 (g) + C2H2 (g)
Now, add equation 3 to the combined equation:
C2H4 (g) + H2 (g) + 2 CH4 (g) -> C2H6 (g) + 3 H2 (g) + C2H2 (g) + C2H4 (g)
Simplify the equation:
2 CH4 (g) -> C2H6 (g) + H2 (g)
The delta H for the desired reaction is the sum of the delta H values from the combined equation:
delta H = -136.4 + 376.8 + (-175.1) = +65.3
Therefore, the correct answer is +65.3. Your calculation is correct.