I need help with this chemistry- enthalpy formation (1 Viewer)

sunshine_c22

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could someone provide me with the answer and working out for this
 

wizzkids

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HINT: remember that the enthalpy of formation of elements in their standard state is zero.
Take the third chemical equation, and write it in reverse.
Multiply the first equation times two.
Multiply the second equation times three.
What do you notice? Can you figure it out?
 

sunshine_c22

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HINT: remember that the enthalpy of formation of elements in their standard state is zero.
Take the third chemical equation, and write it in reverse.
Multiply the first equation times two.
Multiply the second equation times three.
What do you notice? Can you figure it out?
Still confused 😅
 

wizzkids

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HINT:
To derive the enthalpy of formation of C2H5OH (ethanol) we need a chemical equation that has the elements carbon, oxygen and hydrogen on the left-hand side in their standard states, and C2H5OH on the right-hand side. This chemical equation would have zero enthalpy of formation on the left-hand side (because they are all elements) and ethanol on the right-hand side.
 

sunshine_c22

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HINT:
To derive the enthalpy of formation of C2H5OH (ethanol) we need a chemical equation that has the elements carbon, oxygen and hydrogen on the left-hand side in their standard states, and C2H5OH on the right-hand side. This chemical equation would have zero enthalpy of formation on the left-hand side (because they are all elements) and ethanol on the right-hand side.
Is the answer -3004, not sure if its right or wrong. Could you pls provide me with the correct answer if its wrong
 

wizzkids

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Ahh ... no that's not the answer. It is about -284 kJ/mol
equation.png
 

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