31. Carbon monoxide reacts with hydrogen gas to produce a mixture of methane, carbon dioxide, and water. (This mixture is known as substitute natural gas.)
4CO(g) + 8H2(g) → 3CH4(g) + CO2(g) + 2H2O(_)
Use the following thermochemical equations to determine the enthalpy change of the reaction.
C(graphite) + 2H2(g) → CH4(g) + 74.8 kJ
CO(g) + 12O2(g) → CO2(g) + 283.1 kJ
H2(g) + 12 O2(g) → H2O(g) + 241.8 kJ
C(graphite) + 12O2(g) → CO(g) + 110.5 kJ
H2O(l) + 44.0 kJ → H2O(g)
this is required to use the Hess's law, and the manipulation of the equations and mutiplying to get the coeeficients from the orginal question, however i am getting -641kJ...the answer is wrong..any suggestions of what to do..
below is my answer:
multiplied the 1st equation by 3
no change to the second equation
multiplied the 3rd equation by 2
mutiplied the 4th equation by 4 and flipped the equation
fifth equation multiplied by two and flipped
To determine the enthalpy change of the given reaction using Hess's Law, you need to manipulate the given thermochemical equations and multiply them by appropriate coefficients to match the desired overall reaction.
Let's go through the process step-by-step:
1. Start with the given reaction:
4CO(g) + 8H2(g) → 3CH4(g) + CO2(g) + 2H2O(_)
2. Manipulate the equations to match the desired reaction:
a) Multiply the first equation by 3 to match the coefficient of methane:
3[C(graphite) + 2H2(g) → CH4(g)] + 3(74.8 kJ)
3C(graphite) + 6H2(g) → 3CH4(g) + 224.4 kJ
b) The second equation remains unchanged:
CO(g) + 12O2(g) → CO2(g) + 283.1 kJ
c) Multiply the third equation by 2 to match the coefficient of water:
2[H2(g) + 12 O2(g) → H2O(g)] + 2(241.8 kJ)
2H2(g) + 24O2(g) → 2H2O(g) + 483.6 kJ
d) Multiply the fourth equation by 4 (and flip it) to match the coefficient of carbon monoxide:
4[C(graphite) + 12O2(g) → CO(g)] + 4(-110.5 kJ)
4CO(g) → 4C(graphite) + 48O2(g) - 442 kJ
e) Multiply the fifth equation by 2 (and flip it) to match the coefficient of water:
2[H2O(g)] + 2(-44.0 kJ) → 2H2O(l)
2H2O(l) → 4H2O(g) - 88.0 kJ
3. Add up all the manipulated equations:
(3C(graphite) + 6H2(g) → 3CH4(g) + 224.4 kJ)
+(CO(g) + 12O2(g) → CO2(g) + 283.1 kJ)
+(2H2(g) + 24O2(g) → 2H2O(g) + 483.6 kJ)
+(4CO(g) → 4C(graphite) + 48O2(g) - 442 kJ)
+(2H2O(l) → 4H2O(g) - 88.0 kJ)
Simplifying the equation:
3C(graphite) + 6H2(g) + CO(g) + 12O2(g) + 2H2(g) + 24O2(g) + 4CO(g) + 2H2O(l)
→ 3CH4(g) + CO2(g) + 2H2O(g) + 4C(graphite) + 48O2(g)
+ 224.4 kJ + 283.1 kJ + 483.6 kJ - 442 kJ - 88.0 kJ
Cancelling out common terms and simplifying:
3C(graphite) + 6H2(g) + CO(g) + 12O2(g) + 2H2(g)
→ 3CH4(g) + CO2(g) + 2H2O(g) + 4C(graphite) + 48O2(g)
+ 199.1 kJ
The enthalpy change of the reaction is 199.1 kJ.
Please double-check your calculations to ensure you haven't made any mistakes. If you are still obtaining a different answer, please provide the steps you followed and any specific values you used, so we can assist you further.
To solve the problem using Hess's Law, you need to manipulate the given thermochemical equations to obtain the desired reaction equation. Let's go through the steps again to ensure accuracy:
1. Multiply the first equation by 3 to match the coefficient of methane (CH4) in the desired reaction:
3 × (C(graphite) + 2H2(g) → CH4(g) + 74.8 kJ)
= 3C(graphite) + 6H2(g) → 3CH4(g) + 224.4 kJ
2. The second equation remains the same:
CO(g) + 12O2(g) → CO2(g) + 283.1 kJ
3. Multiply the third equation by 2 to adjust the coefficients of water (H2O):
2 × (H2(g) + 12O2(g) → H2O(g) + 241.8 kJ)
= 2H2(g) + 24O2(g) → 2H2O(g) + 483.6 kJ
4. Multiply the fourth equation by 4 (to match the CO coefficient) and flip it:
4 × (C(graphite) + 12O2(g) → CO(g) + 110.5 kJ)
= 4CO(g) + 48O2(g) → 4C(graphite) + 442 kJ
5. Multiply the fifth equation by 2 and flip it to adjust the coefficients of water (H2O):
2 × (H2O(l) + 44.0 kJ → H2O(g))
= 2H2O(g) → 2H2O(l) + 88.0 kJ
Now, add all the manipulated equations together, canceling out the common species on both sides:
3C(graphite) + 6H2(g) + 4CO(g) + 48O2(g) + 2H2O(g) → 3CH4(g) + CO2(g) + 4C(graphite) + 2H2O(l) + 224.4 kJ + 283.1 kJ + 483.6 kJ + 442 kJ + 88.0 kJ
Simplifying the equation:
3C(graphite) + 6H2(g) + 4CO(g) + 48O2(g) + 2H2O(g) → 3CH4(g) + CO2(g) + 4C(graphite) + 2H2O(l) + 1039.1 kJ
Notice that carbon (C) and graphite (C(graphite)) appear on both sides of the equation. Therefore, we can simplify the equation further:
6H2(g) + 4CO(g) + 48O2(g) + 2H2O(g) → 3CH4(g) + CO2(g) + 2H2O(l) + 1039.1 kJ
Now, compare the resulting equation with the given equation:
4CO(g) + 8H2(g) → 3CH4(g) + CO2(g) + 2H2O(_)
From the comparison, we can conclude that the enthalpy change of the reaction is 1039.1 kJ.
It seems that there may have been a calculation error or a mistake in manipulating the equations in your previous attempt. Please double-check the steps and calculations to ensure accuracy.