For an ideal gas, the relation between Cp and Cv is
A Cp − Cv = R
B Cp + Cv = R
C Cp/Cv = R
D Cp × Cv = R
Mayer’s relation for an ideal gas: Cp−Cv=R.
In an isochoric process for a gas, work done is
A positive
B negative
C zero
D maximum
At constant volume, ΔV = 0, so W = −PΔV = 0.
In an isothermal expansion of an ideal gas, internal energy change is
A positive
B negative
C zero
D infinite
For an ideal gas, U depends only on T; isothermal means ΔT = 0 ⇒ ΔU = 0.
For a spontaneous process at constant T and P, the sign of ΔG is
A positive
B negative
C zero
D unpredictable
Spontaneous at constant T,P requires ΔG < 0.
At equilibrium (constant T and P), ΔG is
A positive
B negative
C zero
D maximum
Equilibrium implies no net driving force; ΔG = 0.
If ΔH is negative and ΔS is positive, the reaction is spontaneous
A at low temperature only
B at high temperature only
C at all temperatures
D at no temperature
ΔG = ΔH − TΔS is always negative when ΔH < 0 and ΔS > 0.
The expression relating ΔG°, K at temperature T is
A ΔG° = RT ln K
B ΔG° = −RT ln K
C ΔG° = −RT/K
D ΔG° = RT/K
Fundamental relation between thermodynamics and equilibrium.
If ΔG° = 0, then the value of K is
A 0
B 1
C 10
D infinite
ΔG° = −RT ln K. If ΔG°=0 ⇒ lnK=0 ⇒ K=1.
If K = 10 at a temperature, then ΔG° is
A negative
B positive
C zero
D cannot be predicted
K > 1 implies lnK positive ⇒ ΔG° = −RT lnK negative.
The SI unit of enthalpy change is
A J
B J/mol
C J/mol•K
D atm
Enthalpy change per mole is expressed as J mol⁻¹ (or kJ mol⁻¹).
Entropy change for a reversible process is calculated by
A ΔS = q/T
B ΔS = qrev/T
C ΔS = qirr/T
D ΔS = T/q
Entropy is computed using reversible heat flow for state function evaluation.
Entropy increases when
A gas changes to liquid
B liquid changes to solid
C solid changes to liquid
D gas changes to solid
Disorder increases from solid to liquid; largest increase is liquid to gas.
For an isolated system, a spontaneous process occurs when
A ΔSsystem < 0
B ΔStotal > 0
C ΔH > 0 always
D ΔG > 0 always
Second law: total entropy (system + surroundings) increases for spontaneity.
For reaction 2SO₂ + O₂ ⇌ 2SO₃, Δn (gaseous) is
A −1
B +1
C 0
D −2
Products moles = 2, reactants moles = 3 ⇒ Δn = 2 − 3 = −1.
For a gaseous reaction with Δn = 0, the relation is
A Kp = Kc
B Kp > Kc always
C Kp < Kc always
D Kp = 0
Kp = Kc(RT)^Δn, and (RT)^0 = 1.
If a reaction is reversed, the new equilibrium constant is
A K
B 1/K
C K²
D √K
Reversing swaps numerator/denominator in K expression.
If a balanced reaction is multiplied by 2, the new equilibrium constant becomes
A K/2
B 2K
C K²
D √K
When coefficients are multiplied by n, K becomes Kⁿ.
For an exothermic reaction, increasing temperature causes equilibrium constant to
A increase
B decrease
C become zero
D remain unchanged
For exothermic reactions, heat is like a product; increasing T shifts backward → smaller K.
For an endothermic reaction, increasing temperature causes equilibrium constant to
A increase
B decrease
C become zero
D remain unchanged
Endothermic reactions absorb heat; higher T favors products → K increases.
If Q < K, the reaction proceeds in the direction of
A reactant formation
B product formation
C no change
D both directions equally
System forms more products to raise Q to K.
If Q > K, the reaction proceeds in the direction of
A reactant formation
B product formation
C no change
D catalyst formation
Excess products are converted back to reactants to reduce Q.
Adding an inert gas at constant volume to an equilibrium mixture
A shifts equilibrium forward always
B shifts equilibrium backward always
C does not change equilibrium position
D changes equilibrium constant
At constant volume, partial pressures of reacting gases do not change (only total pressure increases).
Adding an inert gas at constant pressure shifts equilibrium toward side with
A fewer moles of gas
B more moles of gas
C more solids
D more liquids
At constant pressure, volume increases; effectively pressure on reacting gases decreases, favoring more gaseous moles.
pOH of a solution with [OH⁻] = 10⁻⁴ M is
A 2
B 4
C 10
D 14
pOH = −log[OH⁻] = −log(10⁻⁴) = 4.
pH of a solution with [H⁺] = 3.16 × 10⁻⁵ M is approximately
A 4.5
B 5.5
C 3.5
D 6.5
3.16×10⁻⁵ = 10⁻⁴⋅⁵ ⇒ pH = 4.5.
For a conjugate acid-base pair, the relation between Ka and Kb is
A Ka + Kb = Kw
B Ka × Kb = Kw
C Ka/Kb = Kw
D Ka − Kb = Kw
For conjugate pair at 25°C: KaKb = Kw.
The pH of a basic buffer is calculated using
A pH = pKa + log([salt]/[acid])
B pOH = pKb + log([salt]/[base])
C pH = pKb + log([base]/[salt])
D pH = −log Kb
For basic buffer (weak base + its salt): pOH formula is used.
Which combination forms a basic buffer
A NH₄OH and NH₄Cl
B CH₃COOH and CH₃COONa
C HCl and NaCl
D NaOH and NaCl
Weak base + its salt (common ion) forms a basic buffer.
A salt of weak acid and strong base gives solution that is
A acidic
B basic
C neutral
D always buffer
Anion hydrolyzes to produce OH⁻, making solution basic (e.g., CH₃COONa).
A salt of strong acid and weak base gives solution that is
A acidic
B basic
C neutral
D always saturated
Cation hydrolyzes to produce H⁺ (e.g., NH₄Cl).
Which condition is required for precipitation of a sparingly soluble salt
A Qsp < Ksp
B Qsp = Ksp
C Qsp > Ksp
D Qsp = 0
Ionic product greater than Ksp indicates supersaturation → precipitation.
Which salt has the lowest solubility if their Ksp values are: AgCl (10⁻¹⁰), AgBr (10⁻¹³), AgI (10⁻¹⁶)
A AgCl
B AgBr
C AgI
D all equal
Smaller Ksp means less soluble; 10⁻¹⁶ is smallest.
Common ion effect decreases solubility because equilibrium shifts
A forward
B backward
C upward
D randomly
Adding common ion increases product ion concentration, driving dissolution equilibrium back to solid.
If Ksp of AgCl is 1.8×10⁻¹⁰, molar solubility (s) in pure water is closest to
A 1.34×10⁻⁵
B 1.34×10⁻³
C 1.8×10⁻¹⁰
D 9.0×10⁻⁵
For AgCl: Ksp = s² ⇒ s = √(1.8×10⁻¹⁰) ≈ 1.34×10⁻⁵.
If [Ag⁺] = 10⁻³ M, then maximum [Cl⁻] allowed before AgCl precipitates (Ksp = 10⁻¹⁰) is
A 10⁻⁷ M
B 10⁻³ M
C 10⁻¹³ M
D 10⁻⁵ M
Ksp = [Ag⁺][Cl⁻] ⇒ [Cl⁻] = 10⁻¹⁰ / 10⁻³ = 10⁻⁷ M.
In a one-component system along a phase boundary line (two phases), degrees of freedom is
A 0
B 1
C 2
D 3
For C=1, P=2 ⇒ F = 1 − 2 + 2 = 1 (univariant).
At the triple point of a one-component system, degrees of freedom is
A 0
B 1
C 2
D 3
C=1, P=3 ⇒ F=0 (invariant).
The curve representing equilibrium between solid and vapour is
A fusion curve
B sublimation curve
C vaporization curve
D critical curve
Solid ⇌ vapour equilibrium is sublimation line.
The curve representing equilibrium between liquid and vapour is
A fusion curve
B sublimation curve
C vaporization curve
D triple curve
Liquid ⇌ vapour equilibrium is vaporization line.
The curve representing equilibrium between solid and liquid is
A fusion curve
B sublimation curve
C vaporization curve
D critical curve
Solid ⇌ liquid equilibrium is fusion line.
The point where liquid and vapour become indistinguishable is
A triple point
B critical point
C eutectic point
D boiling point
At critical point, boundary between liquid and gas disappears.
In water phase diagram, the fusion curve has negative slope because
A ice is denser than water
B water is denser than ice
C vapour is denser than liquid
D water is an ideal gas
Increasing pressure favors denser phase (liquid), lowering melting point; hence negative slope.
Boiling occurs when vapour pressure becomes equal to
A zero
B internal pressure
C external pressure
D critical pressure
At boiling point, vapour pressure equals external (atmospheric) pressure.
Normal boiling point corresponds to external pressure of
A 0.5 atm
B 1 atm
C 2 atm
D 5 atm
“Normal” boiling point is defined at 1 atm.
Lowering external pressure causes boiling point to
A increase
B decrease
C remain constant
D become infinite
Boiling occurs when vapour pressure equals external pressure; lower external pressure is reached at lower temperature.
For a two-component, one-phase system, degrees of freedom (using phase rule) is
A 1
B 2
C 3
D 0
F = C − P + 2 = 2 − 1 + 2 = 3.
Along any equilibrium line in a binary (two-component) phase diagram with two phases, degrees of freedom is
A 0
B 1
C 2
D 3
F = 2 − 2 + 2 = 2 (bivariant).
At constant temperature, increasing pressure generally favors the phase with
A larger molar volume
B smaller molar volume
C higher entropy always
D higher temperature always
Higher pressure stabilizes the phase occupying less volume (denser phase).
At equilibrium between two phases, chemical potential of a component is
A higher in liquid than vapour
B lower in liquid than vapour
C equal in both phases
D zero in both phases
Phase equilibrium requires equality of chemical potential in the coexisting phases.
A phase diagram represents stability regions in terms of
A pressure and temperature
B volume and mass
C density and viscosity
D entropy and enthalpy
Standard phase diagram plots P vs T showing stable phase regions and equilibrium curves.