Unit 6: Topic 9 - Hess’s Law Flashcards
For a given reaction with some change in enthalpy, explain why reversing the reaction affects the sign of ∆H but not the magnitude.
Let’s use the equation with enthalpy of formation. To calculate the enthalpy change of a reaction, we can sum the enthalpies of formation of the products and subtract the sum of the enthalpies of formation of the reactants. However, if we reverse the reaction, the products become the reactants and the reactants become the products. The formula for the enthalpy change is then negated.
When reversing a chemical reaction, ONLY the sign of ∆H changes.
What does Hess’s law state?
Hess’s Law states that the heat of reaction ∆H for a specific reaction is equal to the sum of the heats of reactions for any set of reactions which in sum equal the original reaction. For example:
C + O2 –> CO2 (1)
2H2 + O2 –> 2H2O (2)
C + 2H2O –> CO2 + 2H2 (3)
if we flip around reaction (2) (let’s call it -2), then (3) is equivalent to (1) + -(2), so the heat of reaction of (3) equals the sum of that of (1) and -(2)
Find ∆H for the reaction PCl5 –> PCl3 + Cl2 given the heat of reactions:
P4 + 6Cl2 –> 4PCl3, ∆H = -2439 kJ
4PCl5 –> P4 + 10Cl2, ∆H = 3438 kJ.
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If we add the two chemical reactions, we have P4 + 4PCl5 + 6Cl2 –> 4PCl3 + P4 + 10Cl2. We can cancel out the P4 and 6Cl2 from both sides (as they are both product and reactant) to get 4PCl5 –> 4PCl3 + 4Cl2. This reaction has heat of reaction -2439 + 3438 = 999 kJ. We have to multiply the reaction by 1/4, so we multiply the heat of reaction by 1/4 as well. 999 * 1/4 = 249.75, so the final heat of reaction is 249.75 kJ.
Container A contains 100 grams of water at 50 degrees Celsius, and container B contains 200 grams of water at 80 degrees Celsius. If the water in both containers is combined, what will be the resulting temperature of the water after a long period of time?
We can use the formula q = mc∆T. Additionally, if two substances at differing temperatures mix, thermal equilibrium is achieved at the temperature for which the heat gain from the colder substance equals the heat loss from the warmer substance. The formula we can use is q1 + q2 = m1c1∆T1 + m2c2∆T2 = 0 (where the 1 is for substance 1 and 2 for substance 2). For water, since c1 = c2, we can simply use 100(∆T1) + 200(∆T2) = 0. Let the final temperature be T. Then ∆T1 = T - 50, and ∆T2 = T - 80. Solving gives T = 70.
