A Level Chemistry Notes: Enthalpy, Kinetics, and Equilibrium

Enthalpy, Rate Equations, and Halogens

Enthalpy Change of Solution

The enthalpy change that accompanies the dissolving of 1 mole of gaseous ions in H₂O to form 1 mol of aqueous ions is -1008/2 = -504 kJ mol^-1. F^- has a more exothermic hydration enthalpy as it has a smaller ionic radius, leading to a stronger attraction to H₂O molecules. The bond enthalpy for F-F is +158 kJ mol^-1.

Rate Equations and Reaction Mechanisms

For the reaction: Fe³⁺ + 2I^- --> FeI₂⁺, the rate equation is: rate = k[Fe³⁺][I^-]², where k = 22.5 dm⁶mol^-2s^-1. The mechanism for this reaction is a two-step process:

1. Fe³⁺ + 2I^- --> FeI₂⁺
2. Fe³⁺ + FeI₂⁺ --> 2Fe²⁺ + I₂

Halogens

Halogens have more electrons, leading to stronger London forces between molecules. This means more heat energy is needed to break these forces, resulting in higher melting points.

Activation Energy and Electrode Potentials

Activation Energy

From the given data, the gradient of the activation energy graph is approximately -910, which corresponds to -E_a/R.

Electrode Potentials

question: - Need a voltmeter and a salt bridge connecting the 2 cells, the Ni/Ni²⁺ cell was a stick of Ni and solution of Ni²⁺, the I₂/I⁻ cell was both in solution with a platinum electrode. - Standard conditions are 298K, 100/101 kPa, conc. of 1moldm⁻³.- Zn has the most negative electrode potential given, so it will be the most likely to undergo oxidation and lose electrons. Hence it reacts with the Cr₂O₃⁻/Cr³ cell to form Cr³⁺, and you had to give the overall equation by joining them. Eqm shifts to the left of the Zn half cell and Eqm shifts to the right of the chromium. - The Cr³⁺ formed will react with leftover/excess Zn, and looking at the second cell equation given, Cr³⁺ would be reduced to Cr²⁺ since the Cr³⁺/Cr²⁺ cell has the less negative cell potential, so is more likely to be reduced. Eqm shifts to the left of the Zn half cell and Egm shifts to the right of the chromium. Kp Forward reaction is exothermic as Kp value decreases so must shift left with high temp as Kp less than 1 so lies on left/reactant side. *Kp = p(NO2)^2 / p(NO)^2 x p(O2) *Kp is only affected by temp. so Kp does not change but will shift until re-establishment.* Doubling pressure - denominator of Kp increases more than numerator as more terms on bottom. So Kp decreases. * SO numerator of Kp increases to restore Kp back to original value and thus eqm shifts to the right. * Where there are fewer gaseous molecules.

C was [Cu(NH₃)₄(H₂O)₂]²⁺ D was CuSO₄ (aq) (blue solution) E was Cu (s) (brown solid)

F was Cu(NO₃)₂•3(H₂O) (hydrated salt)

Maths Questions

Cl2O7 for formula of liquid A.

RbIO3 for compound.

95.9% for NaOH in Bleach.

1.2 mol for Kc

0.477% dissociate

Kp question: when pressure increases without volume increase (isovolumetric contraction), due to the imbalanced Kp fraction, the value of Kp initially decreases, equilibrium shifts to products so Kp gradually increases back to the original value (as only temperature affects position of equilibrium)

enthalpy change when a mole of (gaseous) bonds is broken.

For the accuracy of titration end point, I proposed using starch indicator for a sharp blue/black to colourless change, potentially also allow brown to pale yellow, without indicator.