Lesson 8.2.1
8.2.1 Enthalpy cycles and Hess's Law Quiz: Pearson Edexcel Chemistry, Unit 8
20 questions
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Lesson 8.2.1, Enthalpy cycles and Hess's Law: 20 multiple choice questions for the Pearson Edexcel Chemistry (9CH0), Unit 8: Energetics I, written with Revision Ninja.
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The 20 questions
-
According to Hess's law, enthalpy change is independent of which factor?
- Temperature change
- Final state
- Reaction route
- Initial state
-
Given C(s) + O2(g) -> CO2(g), delta H = -393.5 kJ mol-1, and CO(g) + 1/2 O2(g) -> CO2(g), delta H = -283.0 kJ mol-1, what is delta H for C(s) + 1/2 O2(g) -> CO(g)?
- -110.5 kJ mol-1
- -283.0 kJ mol-1
- +110.5 kJ mol-1
- -676.5 kJ mol-1
-
What must direct and indirect routes share in a Hess cycle?
- Same activation energy
- Same reaction rate
- Same start and end
- Same intermediate steps
-
What happens to the sign of ΔH when a reaction equation is reversed?
- It becomes zero
- It doubles
- It changes sign
- It stays positive
-
What happens to ΔH when all stoichiometry numbers in an equation are doubled?
- It squares
- It halves
- It doubles
- It stays unchanged
-
Using enthalpy of formation data, delta Hf of CO2 = -393.5, H2O(l) = -285.8, and C2H5OH(l) = -277.7 kJ mol-1, what is delta Hc of ethanol? C2H5OH + 3O2 -> 2CO2 + 3H2O
- -1366.7 kJ mol-1
- +1366.7 kJ mol-1
- -956.9 kJ mol-1
- -1088.0 kJ mol-1
-
What is the standard enthalpy change of formation of an element in its standard state?
- -1 kJ mol⁻¹
- +1 kJ mol⁻¹1
- Variable
- Zero
-
How do direct and indirect routes compare in total enthalpy change in a Hess cycle?
- They are equal
- Direct is greater
- They are opposite
- Indirect is greater
-
Given delta H1 = -285.8 kJ mol-1 for H2 + 1/2 O2 -> H2O(l) and delta H2 = -241.8 kJ mol-1 for H2 + 1/2 O2 -> H2O(g), what is delta H for H2O(l) -> H2O(g)?
- +241.8 kJ mol-1
- -527.6 kJ mol-1
- +44.0 kJ mol-1
- -44.0 kJ mol-1
-
What is the enthalpy change for the reaction C(s) + 2H2(g) -> CH4(g), given C(s) + O2 -> CO2 = -393.5, H2 + 1/2 O2 -> H2O(l) = -285.8 and CH4 + 2O2 -> CO2 + 2H2O(l) = -890.3 kJ mol-1? Use Hess's Law.
- -74.8 kJ mol-1
- -1569.6 kJ mol-1
- -890.3 kJ mol-1
- +74.8 kJ mol-1
-
Why do chemists construct Hess cycles for certain chemical reactions?
- Speed up reactions
- Determine reaction rate
- Increase reaction yield
- Calculate unmeasurable enthalpies
-
What must be identical for all routes in a Hess cycle?
- Total entropy change
- Reaction pathway
- Initial and final states
- Activation energy barrier
-
In a Hess cycle using combustion data, where do the enthalpy arrows point?
- To original reactants
- To gaseous elements
- To solid catalysts
- To combustion products
-
Given that the enthalpy of combustion of hydrogen is -285.8 kJ mol-1, how much heat is released when 4.00 g of hydrogen (Mr = 2.0) burns to form liquid water?
- 285.8 kJ
- 571.6 kJ
- 142.9 kJ
- 1143.2 kJ
-
A student calculates a positive ΔH for an exothermic reaction. What error occurred?
- Multiplied by volume
- Incorrect mass unit
- Incorrect sign reversal
- Forgot gas constant
-
What is the standard enthalpy change of formation of oxygen gas, O2(g)?
- 0 kJ mol⁻¹
- +100 kJ mol⁻¹
- +249 kJ mol⁻¹
- -394 kJ mol⁻¹
-
What is the enthalpy change for C(s) + 1/2 O2(g) -> CO(g), given C(s) + O2 -> CO2 = -393.5 and CO + 1/2 O2 -> CO2 = -283.0 kJ mol-1, if the second equation is multiplied by -1?
- -283.0 kJ mol-1
- -110.5 kJ mol-1
- -676.5 kJ mol-1
- +110.5 kJ mol-1
-
What sign does the enthalpy change ΔH have for an exothermic reaction?
- Negative
- Zero
- Positive
- Neutral
-
What is indicated when two Hess cycle routes give conflicting values?
- Endothermic change
- Measurement error
- Activation energy barrier
- Non-standard conditions
-
The standard enthalpy of combustion of carbon is -393.5 kJ mol-1. How much heat is released when 6.00 g of carbon (Ar = 12.0) burns completely?
- 393.5 kJ
- 196.8 kJ
- 98.4 kJ
- 787.0 kJ
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