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NEET-XI-Chemistry

06: Chemical Thermodynamics

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  • #
    Chapter 6 - Chemical Thermodynamics
  • Qstn #1
    Choose the correct answer. A thermodynamic state function is a quantity
    (i) used to determine heat changes
    (ii) whose value is independent of path
    (iii) used to determine pressure volume work
    (iv) whose value depends on temperature only.
    Ans : A thermodynamic state function is a quantity whose value is independent of a path.

    Functions like p, V, T etc. depend only on the state of a system and not on the path.

    Hence, alternative (ii) is correct.
  • Qstn #2
    For the process to occur under adiabatic conditions, the correct condition is:
    (i) ΔT = 0
    (ii) Δp = 0
    (iii) q = 0
    (iv) w = 0
    Ans : A system is said to be under adiabatic conditions if there is no exchange of heat between the system and its surroundings. Hence, under adiabatic conditions, q = 0.

    Therefore, alternative (iii) is correct.
  • Qstn #3
    The enthalpies of all elements in their standard states are:
    (i) unity
    (ii) zero
    (iii) < 0
    (iv) different for each element
    Ans : The enthalpy of all elements in their standard state is zero.

    Therefore, alternative (ii) is correct.
  • Qstn #4
    ΔUθof combustion of methane is - X kJ mol-1. The value of ΔHθ is
    (i) = ΔUθ
    (ii) > ΔUθ
    (iii) < ΔUθ
    (iv) = 0
    Ans : Since ΔHθ = ΔUθ + ΔngRT and ΔUθ = -X kJ mol-1,

    ΔHθ = (-X) + ΔngRT.

    ⇒ ΔHθ < ΔUθ

    Therefore, alternative (iii) is correct.
  • Qstn #5
    The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, -890.3 kJ mol-1 -393.5 kJ mol-1, and -285.8 kJ mol-1 respectively. Enthalpy of formation of CH4(g) will be
    (i) -74.8 kJ mol-1 (ii) -52.27 kJ mol-1

    (iii) +74.8 kJ mol-1 (iv) +52.26 kJ mol-1.
    Ans : According to the question,



    Thus, the desired equation is the one that represents the formation of CH4 (g) i.e.,







    Enthalpy of formation of CH4(g) = -74.8 kJ mol-1

    Hence, alternative (i) is correct.
  • Qstn #6
    A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be
    (i) possible at high temperature
    (ii) possible only at low temperature
    (iii) not possible at any temperature
    (iv) possible at any temperature
    Ans : For a reaction to be spontaneous, ΔG should be negative.

    ΔG = ΔH - TΔS

    According to the question, for the given reaction,

    ΔS = positive

    ΔH = negative (since heat is evolved)

    ⇒ ΔG = negative

    Therefore, the reaction is spontaneous at any temperature.

    Hence, alternative (iv) is correct.
  • Qstn #7
    In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?
    Ans : According to the first law of thermodynamics,

    ΔU = q + W (i)

    Where,

    ΔU = change in internal energy for a process

    q = heat

    W = work

    Given,

    q = + 701 J (Since heat is absorbed)

    W = -394 J (Since work is done by the system)

    Substituting the values in expression (i), we get

    ΔU = 701 J + (-394 J)

    ΔU = 307 J

    Hence, the change in internal energy for the given process is 307 J.
  • Qstn #8
    The reaction of cyanamide, NH2CN(s),with dioxygen was carried out in a bomb calorimeter, and ΔU was found to be –742.7 kJ mol–1at 298 K. Calculate enthalpy change for the reaction at 298 K.

    $$\ce{NH2CN(s) + \frac32 O2(g) -> N2(g) + CO2(g) + H2O(l) } $$
    Ans : Enthalpy change for a reaction (ΔH) is given by the expression,

    ΔH = ΔU + ΔngRT

    Where,

    ΔU = change in internal energy

    Δng = change in number of moles

    For the given reaction,

    Δng = ∑ng (products) – ∑ng (reactants)

    = (2 – 1.5) moles

    Δng = 0.5 moles

    And,

    ΔU = –742.7 kJ mol–1

    T = 298 K

    R = 8.314 × 10–3 kJ mol–1 K–1

    Substituting the values in the expression of ΔH:

    ΔH = (–742.7 kJ mol–1) + (0.5 mol) (298 K) (8.314 × 10–3 kJ mol–1 K–1)

    = –742.7 + 1.2

    ΔH = –741.5 kJ mol–1
  • Qstn #9
    Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol-1 K-1.
    Ans : From the expression of heat (q),

    q = m. c. ΔT

    Where,

    c = molar heat capacity

    m = mass of substance

    ΔT = change in temperature

    Substituting the values in the expression of q:



    q = 1066.7 J

    q = 1.07 kJ
  • Qstn #10
    Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at -10.0°C. ΔfusH = 6.03 kJ mol-1 at 0°C.

    Cp[H2O(l)] = 75.3 J mol-1 K-1

    Cp[H2O(s)] = 36.8 J mol-1 K-1
    Ans : Total enthalpy change involved in the transformation is the sum of the following changes:

    (a) Energy change involved in the transformation of 1 mol of water at 10°C to 1 mol of water at 0°C.

    (b) Energy change involved in the transformation of 1 mol of water at 0° to 1 mol of ice at 0°C.

    (c) Energy change involved in the transformation of 1 mol of ice at 0°C to 1 mol of ice at -10°C.



    = (75.3 J mol-1 K-1) (0 - 10)K + (-6.03 × 103 J mol-1) + (36.8 J mol-1 K-1) (-10 - 0)K

    = -753 J mol-1 - 6030 J mol-1 - 368 J mol-1

    = -7151 J mol-1

    = -7.151 kJ mol-1

    Hence, the enthalpy change involved in the transformation is -7.151 kJ mol-1.
  • Qstn #11
    Enthalpy of combustion of carbon to CO2 is -393.5 kJ mol-1. Calculate the heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas.
    Ans : Formation of CO2 from carbon and dioxygen gas can be represented as:



    (1 mole = 44 g)

    Heat released on formation of 44 g CO2 = -393.5 kJ mol-1

    Heat released on formation of 35.2 g CO2



    = -314.8 kJ mol-1
  • Qstn #12
    Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are -110 kJ mol-1, - 393 kJ mol-1, 81 kJ mol-1 and 9.7 kJ mol-1 respectively. Find the value of ΔrH for the reaction:

    N2O4(g) + 3CO(g)

    N2O(g) + 3CO2(g)
    Ans : ΔrH for a reaction is defined as the difference between ΔfH value of products and ΔfH value of reactants.



    For the given reaction,

    N2O4(g) + 3CO(g)

    N2O(g) + 3CO2(g)



    Substituting the values of ΔfH for N2O, CO2, N2O4, and CO from the question, we get:



    Hence, the value of ΔrH for the reaction is.
  • Qstn #13
    Given

    ; ΔrHθ = -92.4 kJ mol-1

    What is the standard enthalpy of formation of NH3 gas?
    Ans : Standard enthalpy of formation of a compound is the change in enthalpy that takes place during the formation of 1 mole of a substance in its standard form from its constituent elements in their standard state.

    Re-writing the given equation for 1 mole of NH3(g),



    Standard enthalpy of formation of NH3(g)

    = ½ ΔrHθ

    = ½ (-92.4 kJ mol-1)

    = -46.2 kJ mol-1
  • Qstn #14
    Calculate the standard enthalpy of formation of CH3OH(l) from the following data:

    CH3OH(l) + O2(g) CO2(g) + 2H2O(l) ; ΔrHθ = -726 kJ mol-1

    C(g) + O2(g) CO2(g) ; ΔcHθ = -393 kJ mol-1

    H2(g) +O2(g) H2O(l) ; ΔfHθ = -286 kJ mol-1.
    Ans : The reaction that takes place during the formation of CH3OH(l) can be written as:

    C(s) + 2H2O(g) + O2(g)

    CH3OH(l) (1)

    The reaction (1) can be obtained from the given reactions by following the algebraic calculations as:

    Equation (ii) + 2 × equation (iii) - equation (i)

    ΔfHθ [CH3OH(l)] = ΔcHθ + 2ΔfHθ [H2O(l)] - ΔrHθ

    = (-393 kJ mol-1) + 2(-286 kJ mol-1) - (-726 kJ mol-1)

    = (-393 - 572 + 726) kJ mol-1

    ΔfHθ [CH3OH(l)] = -239 kJ mol-1
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