Thermochemistry
Given an endothermic reaction with a positive ΔH value, which scenario would result in an increased magnitude for ΔH if all other conditions remain constant?
Conducting the reaction at higher temperatures
Decreasing the external pressure on the system
Increasing reactant concentrations
Adding a catalyst
Which transition metal would you predict to form a colored compound when bonded with chloride ions based on its electronic configuration?
[Zirconium(Zr)Kr5 s² ,4d²]
[Titanium (Ti), Ar4 s² ,3d² ]
[Zinc (Zn), Ar4 s² ,3d¹⁰ ]
[Scandium (Sc), Ar4 s² ,3d¹ ]
Given that both reactants and products are gases, how would changing pressure influence the sign and magnitude of enthalpy change during a chemical reaction?
Decreasing pressure causes systems absorbing heat becoming more endothermic as they expand.
Changes in pressure induce Le Chatelier's principle shifts yielding varied amounts released/absorbed heats based phase transitions involved.
Increasing pressure decreases volume thus releases heat making reactions more exothermic.
Pressure changes do not affect enthalpy directly since it's a state function independent from work done on/by gas expansion/contraction.
If the temperature of a reaction is increased, what is the most likely effect on the rate of the reaction?
The temperature change will have no effect on the rate.
The rate of the reaction will increase.
The reaction will only proceed if a catalyst is present.
The rate of the reaction will decrease.
Why Might An Atom With A Half-Filled P Subshell Be Less Reactive Than Other Atoms In Its Group?
Half-Filled Subshells Have Symmetrical Electron Distribution Creating Stability That Makes Them Less Likely TO React.
The Presence OF Unpaired Eletroins INa Hlaf-filled P Subshlle Could Increase Repulsion Between AToms Reducing Reactivity.
Atoms With Half-Filled P Loss Sble Shells May Exhibit Lower Electronegativitiy Thus Demonstrating Reduced Tendency TO Forge Newbonds.
An Atom With Half-Filled P Subshell May Have Higher Atomic Mass Making It Less Reactive Due TO Nuclear Shielding.
What outcome might you expect when applying Hess’s Law to a complex thermochemical cycle involving hypothetical intermediates whose formation and decomposition have unknown enthalpies?
It allows determination of overall reaction enthalpy without knowing individual steps' details.
Incorrectly predicts that intermediates with unknown properties do not affect total enthalpy.
Assumes intermediates’ properties can be derived from products and reactants alone.
Accurate prediction requires knowledge of all intermediate steps’ contributions.
In an exothermic reaction, how does the enthalpy of the products compare to the reactants?
It is equal to that of reactants.
It is higher than that of reactants.
Entropy determines this relationship, not enthalpy.
It is lower than that of reactants.

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If the enthalpies of formation for reactants and products are known, how would one calculate ΔH for the reaction using the data?
Dividing the sum of the reactants' enthalpies by the sum of the products' enthalpies
Summing up internal energies for products and subtracting them from internal energies of reactants
Multiplying the difference between the product and reactant enthalpies by the number of moles involved in the reaction
Summing up the enthalpies of formation of products and subtracting the sum of the reactants' enthalpies of formation
If the enthalpy change for a reaction is negative and the entropy change is positive, how will the temperature affect the spontaneity of the reaction?
Spontaneity decreases as temperature increases due to greater enthalpic contributions.
The reaction will be non-spontaneous at all temperatures because enthalpy dominates entropy.
Spontaneity increases as temperature decreases because TΔS becomes less significant.
Spontaneity will not depend on temperature since ΔG would always be negative.
How would you categorize a chemical reaction with a positive ΔH?
Isobaric reaction
Exothermic reaction
Isothermal reaction
Endothermic reaction