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Short chemistry question

2-methylpropene can be obtained by dehydrating 2-methylpropan-1-ol.
(CH3)2CHCH2OH(l) →*(CH3)2C=CH2(g) + H2O(l)
The standard enthalpies of formation of 2-methylpropan-1-ol, 2-methylpropene and water are −335, −17 and −286 kJ mol−1*respectively.
What is the standard enthalpy change for the dehydration of 2-methylpropan-1-ol?

Idk did I done it right, I was thinking since the 2-methylpropan-1-ol did not shows up as a final product, so only add up heat of 2-methylpropan-1-ol and water, is it the right answer?
with these kinds of questions, what do you normally need to do if you are given values for standard enthalpies of formation?

Spoiler

Ty! I tried to calculate it, since formation of water have same enthalpy as these two adds up, so the equation should be -286-(-335-17)=66kJ/mol, is it the right answer?
Original post by bl0bf1sh
with these kinds of questions, what do you normally need to do if you are given values for standard enthalpies of formation?

Spoiler


not quite sure what you mean by "formation of water have same enthalpy as these two adds up"?
I got a different value

have you drawn a Hess cycle? are your arrows pointing in the right direction? are you changing the sign (+/-) if you are going the opposite way to the direction of the arrow?

handy page on Hess cycles: https://chemguide.co.uk/physical/energetics/sums.html
Original post by Lemonadestars
2-methylpropene can be obtained by dehydrating 2-methylpropan-1-ol.
(CH3)2CHCH2OH(l) →*(CH3)2C=CH2(g) + H2O(l)
The standard enthalpies of formation of 2-methylpropan-1-ol, 2-methylpropene and water are −335, −17 and −286 kJ mol−1*respectively.
What is the standard enthalpy change for the dehydration of 2-methylpropan-1-ol?

Idk did I done it right, I was thinking since the 2-methylpropan-1-ol did not shows up as a final product, so only add up heat of 2-methylpropan-1-ol and water, is it the right answer?


The equation shows the dehydration. If yo are using formation enthalpy then the sum of the formation enthalpies of the products minus the enthalpy of formation of the reactant will give you the reaction enthalpy.

Here is a video that shows you how to apply enthalpy changes to determine reaction enthalpy.

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