Physical Chemistry Numericals and Formulas

Flashcards covering formulas and numerical problem-solving techniques in physical chemistry: Solutions, Electrochemistry, and Chemical Kinetics.

How is elevation in boiling point ($\Delta T_b$) calculated using molality, and how is the boiling point of the solution ($T_b$) determined?

$\Delta T_b = K_b \times m$, where $K_b$ is the molal boiling point elevation constant and $m$ is molality. The boiling point of the solution is given by $T_b = T_b^\circ + \Delta T_b$, where $T_b^\circ$ is the boiling point of the pure solvent.

What is the boiling point of a solution prepared by dissolving $18\,\text{g}$ of glucose in $1\,\text{kg}$ of water? (Given: $K_b = 0.52\,\text{K\,kg\,mol}^{-1}$, boiling point of pure water = $373.15\,\text{K}$, molar mass of glucose = $180\,\text{g\,mol}^{-1}$)

$373.202\,\text{K}$. Molality $m = \frac{18\,\text{g} / 180\,\text{g\,mol}^{-1}}{1\,\text{kg}} = 0.1\,\text{mol\,kg}^{-1}$. $\Delta T_b = 0.52 \times 0.1 = 0.052\,\text{K}$. Boiling point $T_b = 373.15 + 0.052 = 373.202\,\text{K}$.

Why is the value of the van 't Hoff factor ($i$) for ethanoic acid in benzene close to $0.5$?

Ethanoic acid ($\text{CH}_3\text{COOH}$) undergoes dimerization (association into double molecules) through hydrogen bonding in non-polar solvents like benzene, halving the total number of particles.

When $25.6\,\text{g}$ of sulphur is dissolved in $1000\,\text{g}$ of benzene, the freezing point is lowered by $0.512\,\text{K}$. What is the molecular formula of sulphur ($\text{S}_x$)? (Given: $K_f = 5.12\,\text{K\,kg\,mol}^{-1}$, atomic mass of sulphur = $32\,\text{g\,mol}^{-1}$)

$\text{S}_8$. Using $\Delta T_f = K_f \cdot \frac{w_B}{M_B \times W_A\,(\text{in kg})}$, $0.512 = 5.12 \times \frac{25.6}{M_B \times 1}$, yielding molar mass $M_B = 256\,\text{g\,mol}^{-1}$. Since $x = \frac{256}{32} = 8$, the formula is $\text{S}_8$.

What is the Nernst equation for an electrode reaction at $298\,\text{K}$?

$$E = E^\circ - \frac{0.0591}{n} \log_{10} \left(\frac{[\text{Products}]}{[\text{Reactants}]}\right)$$

What is the reduction potential of $\text{Fe}^{2+} (0.01\,\text{M}) \mid \text{Fe}(s)$ at room temperature if $E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -0.45\,\text{V}$?

$-0.5091\,\text{V}$. Using $E = E^\circ - \frac{0.0591}{2} \log_{10} \left(\frac{1}{[\text{Fe}^{2+}]}\right) = -0.45 - \frac{0.0591}{2} \log_{10} \left(\frac{1}{0.01}\right) = -0.45 - 0.0591 = -0.5091\,\text{V}$.

How is the degree of dissociation ($\alpha$) of a weak electrolyte related to its molar conductivity?

$\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$, where $\Lambda_m$ is the molar conductivity at a given concentration and $\Lambda_m^\circ$ is the limiting molar conductivity.

The conductivity of $0.01\,\text{M}$ acetic acid is $7.8 \times 10^{-5}\,\text{S\,cm}^{-1}$. What is its degree of dissociation if its limiting molar conductivity is $390\,\text{S\,cm}^2\,\text{mol}^{-1}$?

$\alpha = 0.02$ (or $2\%$). Molar conductivity $\Lambda_m = \frac{\kappa \times 1000}{M} = \frac{7.8 \times 10^{-5} \times 1000}{0.01} = 7.8\,\text{S\,cm}^2\,\text{mol}^{-1}$. Then $\alpha = \frac{7.8}{390} = 0.02$.

What is the formula for standard Gibbs free energy change ($\Delta G^\circ$) in terms of standard cell potential ($E^\circ_{\text{cell}}$)?

$\Delta G^\circ = -nFE^\circ_{\text{cell}}$, where $n$ is the number of moles of electrons transferred and $F$ is Faraday's constant ($96500\,\text{C\,mol}^{-1}$).

What is $\Delta G^\circ$ for the cell reaction $\text{Zn} + \text{Cd}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cd}$ at $25^\circ\text{C}$, given $E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76\,\text{V}$, $E^\circ_{\text{Cd}^{2+}/\text{Cd}} = +0.40\,\text{V}$, and $F = 96500\,\text{C\,mol}^{-1}$?

$-223880\,\text{J\,mol}^{-1}$. $E^\circ_{\text{cell}} = 0.40 - (-0.76) = +1.16\,\text{V}$. $\Delta G^\circ = -2 \times 96500 \times 1.16 = -223880\,\text{J\,mol}^{-1}$.

What is Kohlrausch's law of independent migration of ions?

The limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte ($\Lambda_m^\circ = \nu_+ \lambda_+^\circ + \nu_- \lambda_-^\circ$).

What is the emf of the cell $\text{Zn}(s) \mid \text{Zn}^{2+}(0.01\,\text{M}) \parallel \text{Ag}^+(0.001\,\text{M}) \mid \text{Ag}(s)$, given $E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76\,\text{V}$ and $E^\circ_{\text{Ag}^+/\text{Ag}} = +0.80\,\text{V}$?

$1.4418\,\text{V}$. $E^\circ_{\text{cell}} = 0.80 - (-0.76) = 1.56\,\text{V}$. Applying the Nernst equation: $E_{\text{cell}} = 1.56 - \frac{0.0591}{2} \log_{10} \left(\frac{[\text{Zn}^{2+}]}{[\text{Ag}^+]^2}\right) = 1.56 - \frac{0.0591}{2} \log_{10} \left(\frac{0.01}{(0.001)^2}\right) = 1.56 - 0.1182 = 1.4418\,\text{V}$.

According to Faraday's first law of electrolysis, how is the mass of a substance deposited calculated?

$m = Z \cdot i \cdot t = \frac{\text{Molar Mass}}{n \times 96500} \cdot i \cdot t$, where $i$ is current in amperes, $t$ is time in seconds, and $n$ is the valence factor.

How much copper is deposited when a $5\,\text{A}$ current is passed for $1\,\text{hour}$ through a copper sulphate solution? (Molar mass of $\text{Cu} = 63.5\,\text{g\,mol}^{-1}$)

$5.92\,\text{g}$. Using $m = \frac{63.5 \times 5 \times 3600}{2 \times 96500} \approx 5.92\,\text{g}$.

What are the two essential conditions required for molecular collisions to be effective collisions?

1. Proper orientation of colliding molecules.\n2. Sufficient energy (threshold energy / activation energy) possessed by the reacting molecules.

If the half-life of a zero-order reaction is $2\,\text{min}$, what is the time required for $75\%$ completion of the reaction?

$3\,\text{min}$. For a zero-order reaction, rate constant $k = \frac{[A]_0}{2 t_{1/2}} = \frac{100}{2 \times 2} = 25\,\text{conc\,min}^{-1}$. For $75\%$ completion, remaining concentration $[A] = 25$. Time $t = \frac{[A]_0 - [A]}{k} = \frac{100 - 25}{25} = 3\,\text{min}$.

A first-order reaction is $50\%$ complete in $25\,\text{min}$. What is the time required for $80\%$ completion? (Given: $\log_{10}(5) = 0.6990$)

$58.06\,\text{min}$. Rate constant $k = \frac{0.693}{25} = 0.02772\,\text{min}^{-1}$. For $80\%$ completion, $t = \frac{2.303}{k} \log_{10} \left(\frac{100}{20}\right) = \frac{2.303}{0.02772} \log_{10}(5) \approx 58.06\,\text{min}$.

What is the integrated rate equation to find the rate constant ($k$) of a first-order reaction?

$k = \frac{2.303}{t} \log_{10} \left(\frac{[A]_0}{[A]}\right)$, where $[A]_0$ is the initial concentration and $[A]$ is the concentration at time $t$.

For the reaction $2\text{NO}_2(g) + \text{F}_2(g) \rightarrow 2\text{NO}_2\text{F}(g)$, initial rates are $6 \times 10^{-3}\,\text{M\,min}^{-1}$ at $[\text{NO}_2]=0.2\,\text{M}, [\text{F}_2]=0.05\,\text{M}$; $12 \times 10^{-3}\,\text{M\,min}^{-1}$ at $[\text{NO}_2]=0.4\,\text{M}, [\text{F}_2]=0.05\,\text{M}$; and $4.8 \times 10^{-2}\,\text{M\,min}^{-1}$ at $[\text{NO}_2]=0.8\,\text{M}, [\text{F}_2]=0.10\,\text{M}$. What is the overall order of the reaction?

Overall order is $2$. Comparing experiments 1 and 2 shows the reaction is 1st order with respect to $\text{NO}_2$. Comparing experiments 2 and 3 shows it is 1st order with respect to $\text{F}_2$. Thus, $\text{Rate} = k[\text{NO}_2]^1[\text{F}_2]^1$ and overall order $n = 1 + 1 = 2$.

For a reaction $A + B \rightarrow P$, the rate law is given by $\text{Rate} = k[A]^2 [B]$. How is the reaction rate affected if the concentration of $A$ is doubled, and what is the overall order if $B$ is present in large excess?

(a) The rate increases by $4$ times since \text{Rate} \propto [A]^2$.\n(b) If B$is present in large excess, its concentration remains effectively constant, making the reaction pseudo-second order with an overall order of$2$$.