The pH of neutral water at 25°C is 7.0.

Question: The pH of neutral water at 25C is 7.0. As the temperature increases, ionization of water increases, however, the concentration of H+ ions and OH ions are equal. What will be the pH of pure water at 60C? (i) Equal to 7.0 (ii) Greater than 7.0 (iii) Less than 7.0 (iv) Equal to zero Solution: Option (iii) is the answer....

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When hydrochloric acid is added to cobalt

Question: When hydrochloric acid is added to cobalt nitrate solution at room temperature, the following reaction takes place and the reaction mixture becomes blue. On cooling the mixture it becomes pink. Based on this information mark the correct answer. [Co (H2O)6]3+ (aq) + 4Cl(aq) ⇌ [CoCl4]2 (aq) + 6H2O (l) (pink) (blue) (i) ∆H 0 for the reaction (ii) ∆H 0 for the reaction (iii) ∆H = 0 for the reaction (iv) The sign of ∆H cannot be predicted based on this information. Solution: Option (i) is t...

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Which of the following statements is incorrect?

Question: Which of the following statements is incorrect? (i) In equilibrium mixture of ice and water kept in perfectly insulated flask mass of ice and water does not change with time. (ii) The intensity of red colour increases when oxalic acid is added to a solution containing iron (III) nitrate and potassium thiocyanate. (iii) On addition of the catalyst, the equilibrium constant value is not affected. (iv) The equilibrium constant for a reaction with negative ∆H value decreases as the tempera...

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PCl5, PCl3 and Cl2 are at equilibrium at 500K

Question: PCl5, PCl3 and Cl2 are at equilibrium at 500K in a closed container and their concentrations are 0.8 103 mol L1, 1.2 103 mol L1 and 1.2 103 mol L1 respectively. The value of Kc for the reaction PCl5 (g)  PCl3 (g) + Cl2 (g) will be (i) 1.8 103 mol L1 (ii) 1.8 103 (iii) 1.8 103 L mol1 (iv) 0.55 104 Solution: Option (ii) is the answer....

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Prove that the curves

Question: Prove that the curves $y^{2}=4 x$ and $x^{2}+y^{2}-6 x+1=0$ touch each other at the point $(1,2)$ Solution: Given: Curves $y^{2}=4 x \ldots(1)$ $\ x^{2}+y^{2}-6 x+1=0$ ......(2) $\therefore$ The point of intersection of two curves is $(1,2)$ First curve is $y^{2}=4 x$ Differentiating above w.r.t $\mathrm{x}$, $\Rightarrow 2 y \cdot \frac{d y}{d x}=4$ $\Rightarrow y \cdot \frac{d y}{d x}=2$ $\Rightarrow m_{1}=\frac{2}{y} \ldots(3)$ Second curve is $x^{2}+y^{2}-6 x+1=0$ $\Rightarrow 2 x+...

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Which of the following is not a general

Question: Which of the following is not a general characteristic of equilibria involving physical processes? (i) Equilibrium is possible only in a closed system at a given temperature. (ii) All measurable properties of the system remain constant. (iii) All the physical processes stop at equilibrium. (iv) The opposing processes occur at the same rate and there is dynamic but stable condition. Solution: Option (iii) is the answer....

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For the reaction H2(g) + I2(g) → 2HI (g),

Question: For the reaction H2(g) + I2(g) 2HI (g), the standard free energy is ∆G 0. The equilibrium constant (K ) would be __________. (i) K = 0 (ii) K 1 (iii) K = 1 (iv) K 1 Solution: Option (iv) K 1 is the answer....

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We know that the relationship between Kc and Kp is

Question: We know that the relationship between Kc and Kp is Kp = Kc(RT)∆n What would be the value of ∆n for the reaction NH4Cl (s) NH3 (g) + HCl (g) (i) 1 (ii) 0.5 (iii) 1.5 (iv) 2 Solution: Option (iv)2 is the answer....

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Prove that the curves

Question: Prove that the curves $x y=4$ and $x^{2}+y^{2}=8$ touch each other. Solution: Given: Curves $x y=4$ .....(1) $\ x^{2}+y^{2}=8 \ldots(2)$ Solving $(1) \(2)$, we get $\Rightarrow x y=4$ $\Rightarrow x=\frac{4}{y}$ Substituting $x=\frac{4}{y}$ in $x^{2}+y^{2}=8$, we get, $\Rightarrow\left(\frac{4}{y}\right)^{2}+y^{2}=8$ $\Rightarrow \frac{16}{y^{2}}+y^{2}=8$ $\Rightarrow 16+y^{4}=8 y^{2}$ $\Rightarrow y^{4}-8 y^{2}+16=0$ We will use factorization method to solve the above equation $\Right...

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An ideal gas is allowed to expand against

Question: An ideal gas is allowed to expand against a constant pressure of 2 bar from 10 L to 50 L in one step. Calculate the amount of work done by the gas. If the same expansion were carried out reversibly, will the work is done be higher or lower than the earlier case? (Given that 1 L bar = 100J) Solution: Amount of work done = -pext ∆V = 2 bar (50 10 ) L = 80 L bar. Now, it is given that 1 L bar = 100 J So , -80 L bar = (-80 100) = -8000 J = -8 KJ is the amount of work done in the above proc...

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The enthalpy of reaction for the reaction :

Question: The enthalpy of reaction for the reaction : 2H2(g) + O2(g) 2H2O(l) is ∆HrΘ = 572 kJ mol1 What will be standard enthalpy of formation of H2O (l)? Solution: For the given reaction : 2H2(g) + O2(g) 2H2O(l) the standard enthalpy of reaction is ∆HrΘ = 572 kJ mol1 , so the half of ∆HrΘ will be the standard molar enthalpy of formation ; ΔfHΘ = 1/2 ∆HrΘ = (-572/2) = -286 kJ mol1...

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Show that the curves

Question: Show that the curves $2 x=y^{2}$ and $2 x y=k$ cut at right angles, if $k^{2}=8$ Solution: Given: Curves $2 x=y^{2} \ldots(1)$ $\ 2 x y=k \ldots(2)$ We have to prove that two curves cut at right angles if $k^{2}=8$ Now, Differentiating curves (1) \ (2) w.r.t x, we get $\Rightarrow 2 x=y^{2}$ $\Rightarrow 2=2 y \cdot \frac{d y}{d x}$ $\Rightarrow \frac{d y}{d x}=\frac{1}{y}$ $\mathrm{m}_{1}=\frac{1}{\mathrm{y}} \ldots(3)$ $\Rightarrow 2 x y=k$ Differentiating above w.r.t $x$, $\Rightarr...

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The enthalpy of vapourisation of CCl4

Question: The enthalpy of vapourisation of CCl4 is 30.5 kJ mol1. Calculate the heat required for the vapourisation of 284 g of CCl4 at constant pressure. (Molar mass of CCl4 = 154 g mol1). Solution: The enthalpy of vapourisation is given for 1 mole of CCl4 = 30.5 kJ mol1 Hence, for 284 g, it will be = (mole no. 30.5) kJ Molar mass of CCl4 = 154 g mol1 that means 154 g = 1 mole. Therefore , 284 g = (284g/154gmol-1)= 1.84 mole. Hence the heat required for the vapourisation of 284 g of CCl4 at cons...

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The net enthalpy change of a reaction

Question: The net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus the amount of energy required to form all the bonds in the product molecules. What will be the enthalpy change for the following reaction? H2(g) + Br2(g) 2HBr(g) Given that Bond energy of H2, Br2 and HBr is 435 kJ mol1, 192 kJ mol1 and 368 kJ mol1 respectively. Solution: For the reaction H2(g) + Br2(g) 2HBr(g) Enthalpy change = (Bond energy of H-H bond + Br-Br bond)...

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If the combustion of 1g of graphite produces

Question: If the combustion of 1g of graphite produces 20.7 kJ of heat, what will be molar enthalpy change? Give the significance of the sign also. Solution: The heat of combustion ∆Hc of graphite (i.e. carbon) is given as = 20.7 kJ for 1g of graphite (C). 1 mole of Carbon = 12 g Hence the molar enthalpy change (enthalpy for 1 mole ) = (20.7 12 )= 248.4 KJ mol-1 Since the heat is evolved, the actual molar enthalpy change = -248.4 KJ mol-1 In combustion reactions, heat is always evolved i.e. it i...

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The difference between CP and CV

Question: The difference between CP and CV can be derived using the empirical relation H = U + PV. Calculate the difference between CP and CV for 10 moles of an ideal gas. Solution: For an ideal gas, the difference between these two is CP CV = nR, the universal gas constant and where n= no. of moles Hence for 10 moles of an ideal gas CP CV = 10 R CP CV = 10 8.314 J = 83.14 J...

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Heat capacity (Cp) is an extensive property

Question: Heat capacity (Cp) is an extensive property but specific heat (c) is an intensive property. What will be the relation between Cpand c for 1 mol of water? Solution: 1 mole of water = 18 g. Hence, for water Heat capacity = 18 specific heat. i.e. Cp = 18 C ....

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Show that the curves

Question: Show that the curves $4 x=y^{2}$ and $4 x y=k$ cut at right angles, if $k^{2}=512$. Solution: Given: Curves $4 x=y^{2} \ldots$ (1) $\ 4 x y=k \ldots(2)$ We have to prove that two curves cut at right angles if $k^{2}=512$ Now , Differentiating curves (1) \ (2) w.r.t x, we get $\Rightarrow 4 x=y^{2}$ $\Rightarrow 4=2 y \cdot \frac{d y}{d x}$ $\Rightarrow \frac{d y}{d x}=\frac{2}{y}$ $m_{1}=\frac{2}{y} \ldots(3)$ $\Rightarrow 4 x y=k$ Differentiating above w.r.t $x$, $\Rightarrow 4\left(1...

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Expansion of gas in a vacuum is called

Question: Expansion of gas in a vacuum is called free expansion. Calculate the work is done and the change in internal energy when 1 litre of an ideal gas expands isothermally into a vacuum until its total volume is 5 litre? Solution: Work done in vacuum is calculated by : -w = Text (Vinitial Vfinal ) Pext= 0 w = 0x(5-1) = 0. For isothermal expansion q=o q= ∆U + (-w) Hence ∆U will be 0....

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Although heat is a path function but heats

Question: Although heat is a path function but heats absorbed by the system under certain specific conditions is independent of path. What are those conditions? Explain. Solution: Heat is independent of the path under 2 conditions: 1. When the volume of the system is kept constant- By 1st law of thermodynamics: q = ΔU + (-w) and -w = pΔV Therefore , q = ΔU + pΔV ΔV = 0 (at constant volume) Hence, qv = ΔU + 0 = ΔU= change in internal energy 2. When the pressure of the system is kept constant At c...

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Find the ratio in which the x-axis cuts the join of the points A

Question: Find the ratio in which the x-axis cuts the join of the points A(4, 5) and B(- 10, -2). Also, find the point of intersection. Solution: Let the point which cuts the join of A(4, 5), and B(-10, -2) in the ratio k : 1 be P(x, 0) Formula: If $k: 1$ is the ratio in which the join of two points is divided by another point $(x,$, y), then $\mathrm{x}=\frac{\mathrm{kx}_{2}+\mathrm{x}_{1}}{\mathrm{k}+1}$ $\mathrm{y}=\frac{\mathrm{ky}_{2}+\mathrm{y}_{1}}{\mathrm{k}+1}$ Taking for the y co-ordin...

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Predict the change in internal

Question: Predict the change in internal energy for an isolated system at constant volume. Solution: For an isolated system q=0 and w=0 And according to 1st law of thermodynamics: ∆U= q + w (U=internal energy) Therefore, change in internal energy for an isolated system ∆U = 0....

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Which quantity out of ΔrG

Question: Which quantity out of ΔrG and ΔrGΘwill be zero at equilibrium? Solution: At equilibrium ΔrGΘ = RT lnK . ΔrGΘ can only be 0 when the value of K is =1. So, for all the other K values ΔrGΘ cannot be zero....

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The molar enthalpy of vapourisation

Question: The molar enthalpy of vapourisation of acetone is less than that of water. Why? Solution: Water has strong hydrogen bonds and the high polarity also adds up in resulting it to boil at higher temperatures. Hence water has a higher molar enthalpy than acetone....

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Identify the state functions and path functions

Question: Identify the state functions and path functions out of the following : enthalpy, entropy, heat, temperature, work, free energy. Solution: State functions: enthalpy, entropy, temperature and free energy. Path functions: Heat and work...

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