

Hess's Law and Heat of Reaction
Interactive Video
•
Chemistry, Physics, Science
•
9th - 10th Grade
•
Practice Problem
•
Hard
Patricia Brown
FREE Resource
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10 questions
Show all answers
1.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What is the significance of knowing the state of matter for substances in Hess's Law problems?
It is irrelevant to the calculation.
It changes the chemical formula.
It affects the heat of formation values.
It determines the color of the substance.
2.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Why is it important to balance the chemical equation before calculating the heat of reaction?
To make the equation look neat.
To change the physical state of reactants.
To ensure the correct stoichiometry is used.
To avoid using a calculator.
3.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What is the heat of formation value for oxygen gas, and why?
52.3 kJ/mol, because it is a reactive gas.
0 kJ/mol, because it is in its most stable form.
241.8 kJ/mol, because it is a diatomic molecule.
393.5 kJ/mol, because it is a common reactant.
4.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
How do you calculate the heat of reaction using heats of formation?
By dividing the heats of formation of reactants by products.
By adding the heats of formation of reactants.
By subtracting the sum of heats of formation of reactants from products.
By multiplying the heats of formation of products.
5.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What is the heat of formation for carbon dioxide used in the calculation?
52.3 kJ/mol
393.5 kJ/mol
241.8 kJ/mol
0 kJ/mol
6.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What is the significance of the summation symbol in the heat of reaction calculation?
It indicates multiplication of values.
It represents the total sum of heats of formation.
It shows division of values.
It is used to subtract values.
7.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What does a negative heat of reaction indicate about the nature of the reaction?
The reaction is endothermic.
The reaction is non-spontaneous.
The reaction is exothermic.
The reaction is at equilibrium.
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