Chemical Reaction Equilibria Questions and Answers

Answer: A

Correct answer: expresses every species amount by one variable

With \( n_i = n_{i0} + \nu_i\xi \), every mole number is fixed once the extent is known, so a single reaction has a single unknown however many species take part. That is what makes the equilibrium calculation tractable: the mole fractions are written in terms of \( \xi \) and the equilibrium condition becomes one equation in one variable.

Answer: B

It has units of moles, since it multiplies the dimensionless stoichiometric coefficients to give mole changes. Fractional conversion is the dimensionless quantity it is often confused with. Conversion divides by the initial amount of a reactant.

Answer: C

The relation converts a tabulated thermodynamic quantity into the composition ratio at equilibrium. The negative sign means a negative standard Gibbs energy change gives a constant greater than one. Products are then favoured at equilibrium.

Answer: D

The constant is the exponential of five thousand seven hundred over two thousand five hundred, which is the exponential of about two point three, or roughly ten. A sign slip here inverts the answer. Checking whether the value should exceed one catches it.

Answer: A

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the reason gives the cause. \( K \) is defined by \( \Delta G^{\circ} = -RT\ln K \), and the standard state is at a fixed pressure, so the standard Gibbs energy - and therefore \( K \) - depends only on temperature. Pressure certainly shifts the equilibrium *composition*, but it does so by changing the mole fractions that must satisfy an unchanged \( K \).

Answer: B

A plot of the logarithm of the constant against reciprocal temperature has a slope of minus the standard enthalpy over the gas constant. An exothermic reaction therefore shows a constant that falls as temperature rises.

Answer: C

Correct answer: 1 and 2 only

Statements 1 and 2 are the van ’t Hoff result: \[ \frac{\mathrm{d}\ln K}{\mathrm{d}T} = \frac{\Delta H^{\circ}}{RT^{2}}, \] so \( K \) falls with temperature when the reaction is exothermic and rises when it is endothermic. Statement 3 denies the whole relation, and temperature is in fact the only variable that changes \( K \) at all.

Answer: D

The exothermic reaction wants a low temperature and the reduction in moles wants a high pressure. In practice a compromise temperature is used because the rate at the thermodynamically ideal temperature is far too slow.

Answer: A

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the reason states the compromise. Ammonia synthesis is exothermic, so equilibrium favours low temperature - but at low temperature the rate over the iron catalyst is negligible and the reactor would need to be impossibly large. The plant runs at 400 to 500 degrees, accepting a poorer equilibrium for a workable rate, and recovers the yield by using high pressure and recycling the unconverted gas.

Answer: B

Correct answer: A-2, B-4, C-1, D-3

A reaction producing fewer moles is helped by anything that raises the concentration of the gas. Raising the pressure does so, shifting towards products. An inert added at constant volume changes no partial pressure of a reacting species, so nothing shifts. An inert added at constant *pressure* expands the volume and dilutes everything, which acts like a reduction in pressure and shifts towards reactants. And removing a product continuously drives the reaction forward.