Chapter 9: calorimetry, specific heat equation, ∆H reaction, hess' law

Question 1

How to calculate either specific heat, heat flow (q), or mass or ΔT given the other variables?

Answer 1

Q = mcΔT
- rearrange this formula to solve for ΔT, m, or c
Q = heat (J)
M = mass (g)
C = specific heat capacity (J/g*C)
T = change in temperature (c or k)

Question 2

Equation for standard enthalpy of reaction (ΔH reactions)?

Answer 2

ΔH = [ ΔH products - [ΔH reactants
- find ΔH of products and reactants from table provided and substitute in equation
- the ΔH (enthalpy) of any element in its natural state (ex: Fe2) will be 0 KJ/mol

Question 3

Hess’ law?

Answer 3

ΔH = [ ΔHr
- ΔH = net change in enthalpy
- ΔHr = net change of enthalpy reactions
Manipulate the given enthalpy reactions (by doubling and reversing) to match the target reaction. then find the ΔH of each reaction and add them up to get ΔH of target reaction.
- make sure to double/triple/etc ΔH after doubling/tripling/etc the equation.
- be sure to cancel things that are the same (if you NEED to cancel something, make sure it is on the OPPOSITE side of the equation)
- when reversing the equations, the sign of ΔH will reverse as well. meaning when reveresed + will go to - and - will go to +

Question 4

Example problem: How much energy is released if we react 10 moles of Al?

Answer 4

10 moles Al * -851.5KJ/2 moles Al = -4257.5 KJ
- using ΔH that we found (-851.5KJ), you put that over 2 moles of Al (from molar ratio). This will give us the energy released.

Question 5

Example problem: We want 5,000KJ of energy from this reaction. How many grams of Al do we need to react?

Answer 5

5000 Kj * 2 moles/-851.5Kj * 27.000g Al/1 mol Al = 316 g Al
- take the KJ given and mutiply by the molar ratio over ΔH. Then multiply by gmm (gram molar mass) to get grams of Al.