Heat & Energy
In the equation q = mcΔT, what does c represent?
Specific heat capacity.
Which process increases entropy more:
solid → liquid or liquid → gas?
Liquid → gas.
If ΔG < 0, what does this indicate about a reaction?
The reaction is spontaneous.
What does a rate law describe?
How the reaction rate depends on reactant concentrations.
Which plot produces a straight line for a first-order reaction?
ln[A] vs time.
If a system absorbs 150 J of heat and does 40 J of work on the surroundings, what is ΔU?
ΔU = q + w
ΔU = 150 J + (−40 J)
ΔU = 110 J
(work done by system is negative)
Why does mixing two gases increase entropy even when temperature stays the same?
Mixing gases increases the number of possible molecular arrangements, increasing entropy.
At equilibrium, what is the value of ΔG?
ΔG = 0
If doubling the concentration of A doubles the reaction rate, what is the order in A?
First order.
Which plot produces a straight line for a second-order reaction?
ln[A] vs time.
Why does reversing a chemical reaction change the sign of ΔH?
ΔH depends only on the initial and final states.
Reversing the reaction swaps reactants and products, so the enthalpy change flips sign.
Which change increases entropy the most?
A) freezing water
B) dissolving NaCl in water
C) condensing steam
D) compressing a gas
Correct answer: B – dissolving NaCl in water
Ions disperse in solution, increasing accessible microstates.
If K > 1, what must be true about ΔG°?
ΔG° = −RT lnK
If K > 1, lnK is positive → ΔG° is negative.
If doubling [A] causes the reaction rate to quadruple, what is the reaction order?
Second order.
Why is the half-life of a first-order reaction independent of initial concentration?
First-order half-life depends only on k, not concentration.
Explain why ΔH is a state function but heat q is not.
ΔH is a state function, meaning it depends only on the starting and ending states of the system.
Heat (q) depends on the path taken, so it is not a state function.
Explain why reactions that produce more gas molecules often have positive ΔS.
More gas molecules means more possible molecular positions and motions, increasing entropy.
A reaction has ΔH > 0 and ΔS > 0.
When will this reaction be spontaneous?
Spontaneous at high temperature.
(positive entropy term becomes large enough to overcome ΔH)
Explain why reaction orders cannot be determined from the balanced chemical equation.
Rate laws are determined experimentally and depend on the reaction mechanism, not the overall stoichiometric equation.
How can experimental concentration vs time data be used to determine reaction order?
Plot concentration data using:
[A] vs t
ln[A] vs t
1/[A] vs t
The linear plot reveals reaction order.
Two reactions have identical ΔH values but occur through completely different mechanisms.
Explain why Hess’s Law still allows us to add and manipulate these reactions to determine the overall enthalpy change.
Hess’s Law works because enthalpy is a state function.
Since ΔH depends only on the initial and final states, reactions can be added or rearranged regardless of their mechanisms.
A reaction converts 2 moles of gas into 1 mole of gas but releases a large amount of heat.
Explain how the reaction could still be spontaneous at low temperature but not at high temperature.
If the reaction releases heat, ΔH is negative.
But converting 2 gas → 1 gas gives ΔS negative.
ΔG = ΔH − TΔS
At low temperature, TΔS is small → ΔH dominates → spontaneous.
At high temperature, −TΔS becomes large positive → reaction becomes nonspontaneous.
A reaction has
ΔH = +40 kJ/mol
ΔS = +150 J/mol·K
Explain without calculating numbers whether increasing temperature makes the reaction more or less spontaneous.
ΔS is positive.
As temperature increases, TΔS increases, making ΔG more negative.
So increasing temperature makes the reaction more spontaneous.
For the rate law
rate = k[A]²[B]
If [A] doubles and [B] triples, by what factor does the reaction rate change?
rate = k[A]²[B]
[A] doubles → rate ×4
[B] triples → rate ×3
Total change = 12× increase
A reaction’s half-life doubles when the initial concentration is halved.
Determine the reaction order and explain your reasoning.
This indicates a second-order reaction.
Second-order half-life:
t½ = 1 / (k[A]₀)
When concentration decreases, half-life increases, matching the observation.