General Chemistry III (Chem EH 301) Solved Question 2019: Section III (Physical Chemistry)

Solution to the third Semester Final Examination Question Paper from NEHU, Shillong. 

Paper Chem EH 301, October 2019 Examination.

Section - III (Physical)

Click on Question No to view the solution: [11 (a)] [11 (b)] [11 (c)] [11 (d)]

11(a) Derive an expression for the variation of Gibb's free energy with temperature and pressure. [3]

Solution: The variation of free energy change with variation of temperature and pressure \[G = H - TS\] Since \(H = U + PV\),
Hence, \begin{equation}\label{eq:gibbs}G = U + PV - TS\end{equation}
upon differentiation, \(dG = dU + PdV + VdP - TdS - SdT\)
The First law equation for an infinitesimal change may be written as
\begin{equation*}dq = dU - dw\end{equation*} If the work done is only due to expansion, then, \begin{equation*}\therefore\quad dq = dU + PdV\end{equation*} For a reversible process, \begin{equation}\label{eq:dSrev}dS = \frac{dq}{T}\text{ or, }TdS = dq = dU + PdV\end{equation} Combining equation (\ref{eq:gibbs}) and (\ref{eq:dSrev}), we have, \begin{equation}\label{eq:gibBrev}dG = VdP - SdT\end{equation} This equation gives change of free energy when a system undergoes, reversibly, a change of pressure as well as change of temperature.
If pressure remains constant, i.e., \(dP = 0\), then from equation (\ref{eq:gibBrev}), \begin{equation}dG = -SdT \text{ or, }\left( \frac{\delta G}{\delta T} \right)_P = -S\end{equation} When temperature remains constant, i.e., \(dT = 0\), then, from equation (\ref{eq:gibBrev}) \begin{equation}\label{eq:GibbsT}dG = VdP \text{ or, }\left( \frac{\delta G}{\delta P} \right)_T = V\end{equation} Let the free energy of a system be \(G_1\) in the initial state and \(G_2\) in the final state when an appreciable change in pressure has taken place, at constant temperature. Then, integrating equation (\ref{eq:GibbsT}), the free energy change, \(\Delta G\), is given by \begin{equation*}\Delta G = G_2 - G_1 = \int_{P_1} ^{P_2} VdP \end{equation*} Where, \(P_1\) and \(P_2\) are the initial and final pressures, respectively. if one mole of an ideal gas is under consideration, then \(PV = nRT\). \begin{equation*}\Delta G = RT \int_{P_1} ^{P_2} \frac{dP}{P} = RT\ln \frac{P_2}{P_1} = RT\ln \frac{V_1}{V_2}\end{equation*} Where \(V_1\) and \(V_2\) are the initial and final volumes, respectively.
For n moles of the gas, \begin{equation}\Delta G = nRT\ln \left(\frac{P_2}{P_1}\right) = nRT\ln \left(\frac{V_1}{V_2}\right)\end{equation}

11 (b) Calculate the amount of heat supplied to a Carnot’s cycle working between 368 K and 288 K if the maximum work obtained is 895 joules. [2]

Solution: We know, \[\frac{\text{Work Output}}{\text{Heat Supplied}} = \frac{w}{q} =\text{ efficiency, }\eta = 1 - \frac{T_1}{T_2}, \text{ where, } T_2>T_1\] Given, \(T_1\) = 288K, \(T_2\) = 368 K, Work Output, w = 895 J, q = ?
Therefore, \[\frac{w}{q} = 1 - \frac{T_1}{T_2} = \frac{T_2 - T_1}{T_2}\] \[\text{ or, }q = \frac{w. T_2}{T_2 - T_1} = \frac{895J \times 368K}{368K - 288K} = \frac{329360 J \cancel{K}}{80\cancel{K}} = 4117\ J\]

11 (c) Give the thermodynamic derivation of the law of chemical equilibrium for a general reaction: \(aA + bB + ... \rightleftharpoons lL + mM + ...\) [2.5]

Solution: Consider the general Reversible reaction \[aA + bB + ... \rightleftharpoons lL + mM + ...\] where reactants and the products are assumed to be ideal gas.
We know that chemical potential (i.e., Gibb’s free energy) of reactants consisting of a moles of A and b moles of B is given by the expression: \[G_{reactants} = a\mu_A + b\mu_B\] Where \(\mu_A \text{ and } \mu_B\) are chemical potentials of the species A and B, respectively. Similarly, for the products we have. \begin{equation}\label{eq:Gpro}G_{products} = l\mu_L + m\mu_M\end{equation} In each case, pressure and temperature are constant. The free energy of the reaction is equal to the difference between the free energy of the products and that of the reactants, that is, \begin{align*} (\Delta G)_{reaction} &= G_{products} - G_{reactants}\\ &=(m\mu_M + n\mu_N) - (a\mu_A + b\mu_B) \end{align*} At equilibrium, the free energy change \(\Delta G = 0\) so that equation (\ref{eq:Gpro}) becomes \begin{equation}\label{eq:Gzero}(m\mu_M + n\mu_N) - (a\mu_A + b\mu_B) = 0\end{equation} The chemical potential of the ith species in the gaseous state is given by \begin{equation}\label{eq:Ggas}\mu_i = \mu_i^\ominus + RT\ln\ p_i\end{equation} The chemical \(p_i\) is the partial pressure of the ith component and \(\mu_i^\ominus\) is a standard chemical potential (i.e., when partial pressure of the ith component is unity). From equation (\ref{eq:Gzero}) and (\ref{eq:Ggas}), we obtain: \begin{align*} &&[m(\mu_M^\ominus + RT\ln\ p_M) + n(\mu_N^\ominus + RT\ln\ p_N)]\\ &&- [a(\mu_A^\ominus + RT\ln\ p_A) + b(\mu_B^\ominus + RT\ln\ p_B)] &= 0 \end{align*} \begin{align*} &\text{or,} &RT\ln (p_M^mp_N^n)/(p_A^ap_B^b) &= - [(m\mu_M^\ominus + n\mu_N^\ominus) - (a\mu_A^\ominus - b\mu_B^\ominus)]\\ &&&= - [G_{products} - G_{reactants}] = - (\Delta G)_{reaction}\end{align*} \begin{align}\label{eq:Go}&\text{or,} &(p_M^mp_N^n)/(p_A^ap_B^b)&= e^{\Delta G^\ominus/RT} \end{align} Since \(\Delta G^\ominus\) depends only on temperature and R is the gas constant, hence the right hand side of equation (\ref{eq:Go}) is a constant at constant temperature. Thus, \begin{equation}\label{eq:Kp}(p_M^mp_N^n)/(p_A^ap_B^b) = constant = K_p\end{equation} from equation (\ref{eq:Go}) and (\ref{eq:Kp}) \[K_p = e^{-\Delta G^\ominus/RT} \text{ or } \Delta G^\ominus = - RT\ln K_p\] This equation is known as the van’t Hoff reaction isotherm

11 (d) The equilibrium constant for the reaction \(H_2 (g) + S(g) \rightleftharpoons H_2S (g)\) is 20.2 a.t.m. at 945 ℃ and 9.21 a.t.m. at 1065 ℃. Calculate the heat of the reaction. [2]

Solution: From the integrated van’t Hoff equation. \begin{align*} &&\ln \left(\frac{K_{P,2}}{K_{P,1}}\right) &= \frac{\Delta H^\ominus}{R} \left(\frac{T_2 - T_1}{T_1T_2} \right)\\ &or, &\Delta H^\ominus &= R\ \left(\frac{T_2 - T_1}{T_1T_2} \right)\ln \left(\frac{K_{P,2}}{K_{P,1}}\right)\\ &&&=R\ \left(\frac{T_2 - T_1}{T_1T_2} \right)\ 2.303\log \left(\frac{K_{P,2}}{K_{P,1}}\right) \end{align*} Given, \(K_{P,1}\) = 20.2; \(K_{P,2}\) = 9.21; \(T_1\) = 945 ℃ = 1218 K and \(T_2\) = 1065 ℃ = 1338 K
We know, R = 8.314 \(J\ K^{-1}\ mol^{-1}\) \begin{align*} &\therefore &\Delta H^\ominus &= 8.314 J K^{-1} mol^{-1}\ \left( \frac{1218K \times 1338K}{1338K - 1218K} \right)\ 2.303\log \left(\frac{9.21}{20.2}\right)\\ &&&= 8.314J\cancel{K^{-1}}mol^{-1} \times 13580.7\cancel{K} \times (-0.7855)\\ &&&= -88686.18\ J mol^{-1}\\ &&&= -88.68618\ kJ mol^{-1} \end{align*}


Click on Question No to view the solution: [12 (a)] [12 (b)] [12 (c)] [12 (d)]

12 (a) Differentiate between Gibb’s free energy and Helmholtz free energy. [1.5]

Solution: Gibb’s free energy is represented by the symbol G and Helmholtz free energy is represented by the symbol A. The following reflations hold good for the two quantities. \begin{align*} A = E - TS\\ G= H- TS \end{align*} \(\Delta A\) gives the maximum work obtainable from a system whereas \(\Delta G\) gives the maximum useful work (other than that of expansion) obtainable from the system.

12 (b) One mole of an ideal mono-atomic gas expands reversibly from a volume of 10 dm3 at a temperature of 298K to a volume of 20 dm3 at a temperature of 250K. Assuming that \(C_V = \frac{3}{2}R\), calculate the entropy change for the process. [3]

Solution: Given, T1 = 298K, T2 = 250K, V1 = 10 dm3, V2 = 20 dm3, \(C_V = \frac{3}{2}R\)
We know, R = 8.314\(\ J\ K^{-1}\ mol^{-1}\) \begin{align*} &\text{Also,} &\Delta S &= C_v\ln(T_2/T_1) + R\ln (V_2/V_1)\\ &&&= \frac{3}{2}\ 8.314 J K^{-1} mol^{-1}\times \ln \left(\frac{250K}{298K}\right) +\\ &&& 8.314 J K^{-1} mol^{-1} \times \ln \left(\frac{20dm^3}{10dm^3}\right)\\ &&&= 12.471\ J K^{-1} mol^{-1} \times (-0.175633) + 5.763\ J K^{-1} mol^{-1}\\ &&&= -2.19\ J K^{-1} mol^{-1} + 5.763\ J K^{-1} mol^{-1}\\ &&&= 3.573\ J K^{-1} mol^{-1} \end{align*}

12 (c) Derive van’t Hoff equation for temparature dependence of equilibrium constant KP. [3]

Solution: Van’t Hoff equation for the temperature - dependence of Equilibrium Constant: The equation for reaction isotherm when the reactants as well as the products are gaseous and are also in their standard states, is represente as \begin{equation}\label{eq:VHeq} \Delta G^\ominus = - RT\ln K_p\\ \end{equation} differentiating with respect to temperature, T, at constant pressure we get \begin{equation*} \left[ \frac{\partial(\Delta G^\ominus)}{\partial T} \right]_P = -R\ln K_p - RT\ \frac{d (\ln K_p)}{dT} \end{equation*} multiplying throughout by T, we get, \begin{equation}\label{eq:VHeqT} T\left[ \frac{\partial(\Delta G^\ominus)}{\partial T} \right]_P = -RT\ln K_p - RT^2\ \frac{d (\ln K_p)}{dT} \end{equation} substituting for \ref{eq:VHeq} and \ref{eq:VHeqT}, we have \begin{equation*} T\left[ \frac{\partial(\Delta G^\ominus)}{\partial T} \right]_P = \Delta G^\ominus - RT^2\ \frac{d (\ln K_p)}{dT} \end{equation*} \begin{equation}\label{eq:DG} \text{ or, }\Delta G^\ominus = T\left[ \frac{\partial(\Delta G^\ominus)}{\partial T} \right]_P + RT^2\ \frac{d (\ln K_p)}{dT} \end{equation} From the Gibbs-Helmholtz equation, \begin{equation}\label{eq:GHeq} \Delta G^\ominus = \Delta H^\ominus + T\left[ \frac{\partial(\Delta G^\ominus)}{\partial T} \right]_P \end{equation} From equation \ref{eq:DG} and \ref{eq:GHeq} we have, \[RT^2\ \frac{d(\ln K_p)}{dT} = \Delta H^\ominus\] \begin{equation}\label{eq:vHeq} \text{ or, }\frac{d(\ln K_p)}{dT} = \frac{\Delta H^\ominus}{RT^2} \end{equation} Equation (\ref{eq:vHeq}) is known as the van’t Hoff’s equation. \(\Delta H^\ominus\) is the enthalpy change for the reaction at constant pressure when the reactants as well as the products are in their standard states.
If, over a short temperature range, \(\Delta H^\ominus\) is assumed to be temperature-independent, then on integrating equation (\ref{eq:vHeq}) between temperatures T1 and T2 at which the equilibrium constants \(K_{P,1}\) and \(K_{P,2}\), respectively and assuming that \(\Delta H^\ominus\) remains constant over this range of temperature, we get, \begin{align*} &&\int_{K_{P,1}}^{K_{P,2}} d(\ln K_P) &= \frac{\Delta H^\ominus}{R} \int_{T_1}^{T_2} \frac{dT}{T^2}\\ &\text{or,} &\ln K_{P,2} - \ln K_{P,1} &= \frac{\Delta H^\ominus}{R}\left[\frac{1}{T_1} - \frac{1}{T_2} \right]\\ &&&=\frac{\Delta H^\ominus}{R}\left[\frac{T_2 - T_1}{T_1T_2} \right]\\ \end{align*} \begin{align}\label{eq:ivHeq} &\text{or,} &\log\left(\frac{K_{P,2}}{K_{P,1}} \right) &= \frac{\Delta H^\ominus}{2.303 \times R}\left[\frac{T_2 - T_1}{T_1T_2} \right]\\ \end{align} Equation (\ref{eq:ivHeq}) is the integrated van’t Hoff equation

12 (d) At 473K the equilibrium constant KC for the decomposition of PCl5 is 8.3 x 10-3. If the decomposition proceeds as \({PCl_5 \rightarrow PCl_3 + Cl_2}\), find KP for the reaction. [2]

Solution: Given, T = 473 K, Kc = 8.3 x 10-3 dm-3 mol-1
for the given reaction, \(PCl_5 \rightarrow PCl_3 + Cl_2, \ \Delta n = 2 - 1 = 1\)
We know, R = 0.08206 = dm-3 atm K-1 mol-1 \begin{align*} &\text{Also,}&K_p &= K_c (RT)^{\Delta n}\\ &&&= 8.3 \times 10^{-3}\cancel{dm^3mol^{-1}} \times (0.08206\cancel{dm^3}atm\cancel{K^{-1}mol^{-1}} \times 473\cancel{K})^1\\ &&&=322.15 \times 10^{-3} = 0.32215\ atm \end{align*} Therefore KP = 0.32215 atm


Click on Question No to view the solution: [13 (a)] [13 (b)] [13 (c)] [13 (d)]

13 (a) Derive the integrated rate law for a second-order reaction having one reacting species. [3]

Solution: The second order reaction for a reaction having only one reacting species can be represented as follow \[2A \rightarrow P\] the rate of the reaction would be expressed as \begin{equation}\text{Rate} = \frac {dx}{dt} = k_2(a-x)^2\end{equation} where, a is the initial concentration of A, x is the concentration of the product formed after time t and (a — x) is the concentration of A remaining after time t.
Separating the variables and integrating, we have \begin{equation}\int \frac{dx}{(a-x)^2} = k_2 dt\end{equation} \begin{equation}\text{ or, }\left[-\frac{1}{a-x}\right](-1) = k_2t + C \text{ or, } \frac{1}{a-x} = k_2t + C\end{equation} We know that at \(t = 0\text{, }x = 0\text{, so that }C = \frac{1}{a}\). Hence, \[\frac{1}{a-x} = k_2t + \frac{1}{a}\] \begin{equation}\label{eq:soR}k_2 = \frac{1}{t}\left[ \frac{1}{a-x}-\frac{1}{a}\right] = \frac{1}{t}\left[ \frac{x}{a(a-x)}\right]\end{equation} Equation (\ref{eq:soR}) is the required integrated expression for the rate constant of a second-order reaction in which two molecules of the same reactant are involved in the reaction.

13 (b) Why is the hydrolysis of an ester \(CH_3-COOCH_3 + H_2O\) \(\rightarrow\) \(CH_3COOH + CH_3OH\) called a pseudounimolecular reaction though more than one kind of reactants are involved? [1.5]

Solution: The hydrolysis of an ester \(CH_3-COOCH_3 + H_2O\) \(\rightarrow\) \(CH_3COOH + CH_3OH\) called a pseudounimolecular reaction even though more than one kind of reactants are involved, is because water, H2O is present in excess, so the concentration of water does not change during the reaction. Therefore the rate of the reaction depends only on the concentration of CH3COOCH3. \[\text{Rate } \propto [CH_3-COOCH_3][H_2O]\] Here, \([CH_3-COOCH_3]\) is the concentration of \(CH_3-COOCH_3\) and \([H_2O]\) is the concentration of \(H_2O\). \[\text{Rate } = k_1 [CH_3-COOCH_3] [H_2O]\] \[\text{Rate } = k_2 [CH_3-COOCH_3]\] Where, \(k_2 = k_1 [H_2O]\) \[\therefore \text{Rate } \propto [CH_3-COOCH_3]\]

13 (c) What are colligative properties? Give examples. [2]

Solution: Colligative properties are those properties of solutions that depend on the ratio of the number of solute particles to the number of solvent molecules in a solution, and not on the nature of the chemical species present.
Examples, depression of freezing point, elevation of boiling point etc.

13 (d) A 0.5% aqueous solution of potassium chloride was found to freeze at -0.24℃. Calculate the van’t Hoff factor. [Given Kf = 1.86 K kg mol-1] [3]

Solution: Given,
Kf = 1.86 K kg mol-1, \(\Delta T\) = 0.24K
weight of solute, \(w_2 = 0.5 g = 0.5 \times 10^{-3} kg\)
weight of solvent, \(w_1 = 100 g = 100 \times 10^{-3} kg\)
We know,
Molecular weight of KCl, \(M_{normal}\) = 74.5 g/mol
Observed Molecular weight,
\begin{align*} M_{observed} = \frac{K_fw_2}{w_1\Delta T} = \frac{ (1.86\ K\ kg\ mol^{-1})\ ( 0.5 \times 10^{-3}\ kg)}{(100 \times 10^{-3}\ kg)\ (0.24\ K) }\\ = 0.03875\ kg\ mol^{-1} = 38.75\ g\ mol^{-1} \end{align*} \begin{align*} \text{also, van’t Hoff factor, }i = \frac{M_{normal}}{M_{observed}} = \frac{74.5\ gmol^{-1}}{38.75\ gmol^{-1}} = 1.92 \end{align*}


Click on Question No to view the solution: [14 (a)] [14 (b)] [14 (c)] [14 (d)]

14 (a) From the following data, show that the decomposition of hydrogen peroxide in aqueous solution is a first - order reaction:

Time (min): 0 15 30
V(mL) : 25.4 9.83 3.81

Where V is the volume of potassium permanganate required to decompose a definite volume of hydrogen peroxide. [2]

Solution: Given,

Time (min): 0 15 30
V(mL) : 25.4 9.83 3.81

for a first - order reaction, the rate constant is given as \[k_1 = \frac{1}{t}\ln\frac{a}{a-x}\] Where a is the concentration at time t0, and x is the concentration at a given time tx
From the given data at time, to, a = 25.4, thus,

time (t) a a - x ln a/(a - x) \[k_1 = \frac{1}{t}\ln\frac{a}{a-x}\]
15 25.4 9.83 0.9493 0.0633
30 25.4 3.81 1.8971 0.0632

Since the value of k1 is constant, hence, the reaction is a first-order reaction

14 (b) The rate constant for the first-order reaction is \[\log K(\text{in }s^{-1}) = 14.34 - \frac{1.25 \times 10^4 K}{T}\] calculate the - (i) activation energy; (ii) rate constant K at 427°C [2.5]

SolutionGiven \begin{equation}\label{eq:Given}\log K(\text{in }s^{-1}) = 14.34 - \frac{1.25 \times 10^4 K}{T}\end{equation} (1) We know, \begin{align*} &&k &= Ae^{-E_a/RT}\end{align*} \begin{align}\label{eq:no1} &\text{or,} &\log k &= \log A - \frac{E_a}{2.303\ RT} \end{align} from equation (\ref{eq:Given}) and (\ref{eq:no1}), \begin{align*} &&\frac{E_a}{2.303\ RT} &= \frac{1.25 \times 10^4K}{T}\\ &\text{or,} &E_a &= 2.303 \times R\times 1.25 \times 10^4K\\ &&&=2.303 \times 8.314 J K^{-1} mol^{-1}\times 1.25 \times 10^4K\\ &&&=23.93\times10^4J\ mol^{-1} = 239.3 kJ mol^{-1} \end{align*} (ii) Given, T = 427 °C = 700 K, from equation (\ref{eq:Given}) we get, \begin{align*} &&\log K &= 14.34 - \frac{1.25 \times 10^4 K}{700K}\\ &&&=14.34 - 17.857 = -3.517\\ &\text{or,} &K &= 3.0409 \times 10^{-4} \end{align*} Therefore, the activation energy is 239.3 kJ mol-1 and the rate constant at 427 °C is \(3.0409 \times 10^{-4}\).

14 (c) What is reverse osmosis? What are its applications? [2]


Figure of Reveres Osmosis Process

Reverse osmosis is the phenomenon where the solvent from a solution with higher concentration move to a solution with lower concentration separated by a semi-permeable membrane, when pressure is applied. It is widely used in the purification of water, desalination of sea water etc.

14 (d) The freezing point of a solution containing 0.3g of an organic compound in 30.0g benzene is lowered by 0.45°C. Calculate the molecular weight of the compound. [Given Kf for benzene = 5.12 K kg mol-1] [3]

Solution: Given \begin{align*} &&\text{weight of solute, } w_1 &= 0.3 g, &\Delta T_f &= 0.45K, \\ &&\text{weight of solvent, }w_2 &= 30g = 30/1000\ kg, &K_f &= 5.12 K\ kg\ mol^{-1} \end{align*} We know, \begin{align*} &\text{Molecular weight,} &M &= \frac{K_f \times w_1}{\Delta T_f \times w_2}\\ &\text{or} &M &=\frac{5.12 K\ kg\ mol^{-1} \times 1000 \times 0.3g}{0.45K \times 30 kg}\\ &\text{or,} &M &=\frac{1536}{13.5} g\ mol^{-1} = 113.78 g\ mol^{-1} \end{align*} Therefore, the molecular weight of the organic compound is 113.79 g/mol-1

If you have any queries, please leave a comment below.



F Diengdoh.com: General Chemistry III (Chem EH 301) Solved Question 2019: Section III (Physical Chemistry)
General Chemistry III (Chem EH 301) Solved Question 2019: Section III (Physical Chemistry)
Solution to the third Semester Final Examination Question Paper from NEHU, Shillong. Paper Chem EH 301, October 2019 Examination
F Diengdoh.com
Loaded All Posts Not found any posts VIEW ALL Readmore Reply Cancel reply Delete By Home PAGES POSTS View All RECOMMENDED FOR YOU LABEL ARCHIVE SEARCH ALL POSTS Not found any post match with your request Back Home Sunday Monday Tuesday Wednesday Thursday Friday Saturday Sun Mon Tue Wed Thu Fri Sat January February March April May June July August September October November December Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec just now 1 minute ago $$1$$ minutes ago 1 hour ago $$1$$ hours ago Yesterday $$1$$ days ago $$1$$ weeks ago more than 5 weeks ago Followers Follow THIS PREMIUM CONTENT IS LOCKED STEP 1: Share to a social network STEP 2: Click the link on your social network Copy All Code Select All Code All codes were copied to your clipboard Can not copy the codes / texts, please press [CTRL]+[C] (or CMD+C with Mac) to copy Table of Content