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## Chemistry library

### Course: Chemistry library>Unit 15

Lesson 2: Enthalpy

# Worked example: Using Hess's law to calculate enthalpy of reaction

AP.Chem:
ENE‑3 (EU)
,
ENE‑3.C (LO)
,
ENE‑3.C.1 (EK)
,
ENE‑3.D (LO)
Hess's law can be used to calculate enthalpy changes that are difficult to measure directly. In this video, we'll use Hess's law to calculate the enthalpy change for the formation of methane, CH₄, from solid carbon and hydrogen gas, a reaction that occurs too slowly to be measured in the laboratory. Created by Sal Khan.

## Want to join the conversation?

• I am confused as to why, in the last equation, Sal takes the sum of all of the Delta-H reactions, rather than (Products - Reactants). Could someone please explain to me why this is different to the previous video on Hess's law and reaction enthalpy change.
• You can only use the (products - reactants) formula when you're dealing exclusively with enthalpies of formation. Here, you have reaction enthalpies, not enthalpies of formation, so cannot apply the formula.
• How do you know what reactant to use if there are multiple?
• If you are confused or get stuck about which reactant to use, try to use the equation derived in the previous video (Hess law and reaction enthalpy change). From the given data look for the equation which encompasses all reactants and products, then apply the formula. In this example it would be equation 3.
• If C + 2H2 --> CH4 why is the last equation for Hess's Law not ΔHr = ΔHfCH4 -ΔHfC - ΔHfH2 like in the previous videos, in which case you'd get ΔHr = (890.3) - (-393.5) - (-571.6) = 1855.4
• The equation for the heat of formation is the third equation, and ΔHr = ΔHfCH₄ -ΔHfC - 2ΔHfH₂ = ΔHfCH₄ - 0 – 0 = ΔHfCH₄.

890.3 is NOT the heat of formation of CH₄; it is the reverse of the combustion of CH₄.
-393.5 is NOT the heat of formation of C; it is the reverse of the combustion of C.
-571.6 is NOT the heat of formation of H₂; it is the heat of combustion of H₂
If you add all the heats in the video, you get the value of ΔHCH₄.
• Determine the standard enthalpy change for the formation of liquid hexane (C6H14) from solid carbon (C) and hydrogen gas (H2) from the following data :
C(s) + O2(g) → CO2(g) ΔHAo = -394.4 kJ (reaction A)
2H2(g) + O2(g) → 2H2O(l) ΔHBo = -571.6 kJ (reaction B)
2C6H14(l) + 19O2(g) → 12CO2(g) + 14H2O(l) ΔHCo = -4163.0 kJ (reaction C)
You must write your answer in kJ mol-1 (i.e kJ per mol of hexane).
• Why can't the enthalpy change for some reactions be measured in the laboratory?Which equipments we use to measure it?
• Simply because we can't always carry out the reactions in the laboratory.
For example, CO is formed by the combustion of C in a limited amount of oxygen.
But the reaction always gives a mixture of CO and CO₂.
However, we can burn C and CO completely to CO₂ in excess oxygen.
We can get the value for CO by taking the difference.
• where exactly did you get the other 3 equations to find the first equation?
• What happens if you don't have the enthalpies of Equations 1-3?
• Except you always do. Or you look it up in a source book.
• From the three equations above, how do you know which equation is to be reversed?
• how would we calculate the absorbed heat by sodium acetate crystal plus water while in the microwave? When the specific heat of sodium acetate and water is 0.668 and 1cal'g
• It seems like your question is a heat transfer problem considering your included information about the specific heat. The formula to be used is: Q = mCΔT, where Q is heat transferred, m is mass, C is specific heat, and ΔT is the change in temperature. So you need information about the mass and temperature changes of the water and sodium acetate in addition to the specific heats to be able to solve this problem.
• hi,
I don't understand something about+ and -.
at the end of the video, when you are calculating the total ∆H.
if we know that ∆H= sum of ∆H products - sum of ∆H reactants
shouldn't it be
∆H= 890.3 -(-965.1) = 890.3 + 965 = 1185.4
instead of 890.3 + 965.1 = -74.8
?