Hess's Law
q=mC∆T
Exothermic or Enothermic
Calorimetry
Heat and Enthalpy Changes
100

Define Hess's Law

If a series of reactions are added together, the enthalpy change for the net reaction will be the sum of the enthalpy change for the individual steps.

100
What are the units for C(specific heat)
J/(g*°C)
100
If a reaction gives off energy it is considered what?
Exothermic
100
Is C in q=mC∆T always going to be 4.184 J/(g*°C)?
No the specific heat of water is 4.184 J/(g*°C) but every substance has its own specific heat
100
How much heat will be released when 5.73 g of sulfur reacts with excess O₂ according to the following equation? 2S + 3O₂ → 2SO₃ ∆H = -791.4 kJ
q = 5.73 g S * (1 mol S / 32 g S) * (-791.4 kJ / 2 mol S) q = -70.9 kJ = -70900 J
200

What are the 2 rules for Hess's Law?

1) If you multiply the coefficients by a number you multiply the ΔH by the same number. 

2) If a reaction is reversed, the sign of ΔH is also reversed.

200
What is the specific heat of water?
4.184 J/(g*°C)
200
If the ΔH is positive, what is the reaction?
Endothermic
200
What is the specific heat of silicon if the temperature of a 6.35 g sample of silicon is increased by 5.92 °C when 16.8 J of heat is added?
C = 0.447 J/(g*°C)
200
How much heat will be absorbed when 25.6 g of bromine reacts with excess H₂ according to the following equation? H₂ + Br₂ → 2HBr ∆H = 72.80 kJ
q = 25.6 g Br₂ * (1 mol Br₂ / 159.0 g Br₂) * (72.80 kJ / 1 mol Br₂) q = 11.7 kJ = 11700 J
300

Use the thermochemical equations shown below to determine the enthalpy for the reaction:
H₂O + CO₂ → H₂CO₃
H₂CO + O₂ → H₂CO₃ ΔH= -140kJ
H₂O + CO₂ → H₂CO + O₂ ΔH= 62.5kJ

H₂O + CO₂ → H₂CO₃ ΔH= -77.5kJ

300
Define the 4 parts in q=mC∆T q is? m is? C is? ∆T?
q is the energy that is went into or out of the system. m is the mass of the system C is the specific heat of the system ∆T is the change in temperature of the system
300
When the net energy released is negative
Exothermic
300
What is the specific heat of gold if the temperature of a 6.34 g sample of gold is increased by 14.3 °C when 13.1 J of heat is added?
C = 0.144 J/(g*°C)
300

How much heat will be released when 17.53 g of iron reacts with excess O₂ according to the following equation?
3Fe + 2O₂  →  Fe₃O₄   ∆H = -1120.48 kJ

q = 17.53 g Fe * (1 mol Fe / 55.85 g Fe) * (-1120.48 kJ / 3 mol Fe)
q = -117.2 kJ 

400

Use the thermochemical equations shown below to determine the enthalpy for the reaction:
2C₂H₄O + 2H₂O → 2C₂H₆O + O₂
2CO₂ + 3H₂O → C₂H₆O + 3O₂ ΔH= 1713.7kJ
2CO₂ + 2H₂O → C₂H₄O + 5/2O₂ ΔH= 1458.7kJ

2C₂H₄O + 2H₂O → 2C₂H₆O + O₂ ΔH= 510.0kJ

400
When 2.1 g of chocolate are burnt under a can of 150 g of water. If there are 12 Cal in this much chocolate and the specific heat of water is 4.184 J/gºC. How much would the temperature of the water in the can change?
∆T = 80ºC
400
Energy Diagram
Exothermic if the reactants side is high then the products side. Endothermic if the reactants side is lower then the products side.
400
When a 25.4 g sample of KCN dissolves in 50 g of water in a calorimeter, the temperature drops from 36.2 °C to 14.9 °C. Calculate ∆H for the process. KCN(s) → K⁺(aq) + CN⁻(aq) ∆H = ? Assume that C (of the solution)= 3.34 J/(g*°C)
q =m * C * ∆T q = 5364 J ∆H = q / mol of KCN 25.4 g of KCN * (1 mol of KCN/65 g KCN) = 0.391 mol ∆H = 13700 J/mol
400

How much heat will be absorbed when 21.6 g of nitrogen reacts with excess O₂ according to the following equation?
N₂ + O₂  →  2NO  ∆H = 180 kJ

q = 21.6 g N₂ * (1 mol N₂ / 28.0 g N₂) * (180 kJ / 1 mol N₂)
q = 138.9 kJ

500
Use the thermochemical equations shown below to determine the enthalpy for the reaction: H₂O → H₂ + ½O₂ C₃H₈ + 5O₂ → 3CO₂ + 4H₂O ΔH= -4440kJ C + O₂ → C₂O ΔH= -788kJ 3C + 4H₂ → C₃H₈ ΔH= -208kJ
H₂O → H₂ + ½O₂ ΔH= 571kJ
500
When burning food under a can of water with 200mL of water in the can. The temperature raises from 20ºC to 31ºC. How much energy entered the system in Joules?
q = 9205J
500
In an Endothermic reaction where does the energy for the reaction come from.
Its surroundings Example a cold pack is an endothermic reaction that pulls the energy from the surrounding environment into the cold pack. Because energy must be added to the system(The cold pack) the reaction is endothermic.
500
When a 14.23 g sample of KCl dissolves in 90 g of water in a calorimeter, the temperature drops from 35.5 °C to 23.7 °C. Calculate the ∆H for the process. KCl(s) → K⁺(aq) + Cl⁻(aq) ∆H = ? Assume that C = 3.34 J/(g*°C)
q =m * C * ∆T q = 4108 J ∆H = q / mol of KCl 14.23 g of KCl * (1 mol of KCl/ 75.45 g KCl) = 0.1886 mol ∆H = 21780 J/mol
500

How much heat will be released when 14.6 g of hydrogen reacts with excess O₂ according to the following equation?
2H₂ + O₂  →  2H₂O  ∆H = - 571.6 kJ

q = 14.6 g H₂ * (1 mol H₂ / 2.02 g H₂) * (-571.6 kJ / 2 mol H₂)
q = -2065.7 kJ 

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