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Chemistry Test 205

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Chemistry Test 205
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Weekly Quiz Competition
  • Question 1
    4 / -1

    What is the hybridization shown by C1 and C2 carbons, respectively in the given compound?

    OHC − CH = CH − C2H2C1OOCH3

    Solution

    Ester group has more priority than aldehyde. So numbering should be done from left to right. C1 has double bond and is sp2 hybridised.

     

  • Question 2
    4 / -1

    For irreversible expansion of an ideal gas under isothermal condition, the correct option is:

    Solution
    • For a spontaneous process, ΔStotal > 0 and since irreversible process is always spontaneous therefore ΔStotal > 0.
    • Since ΔU = nCvΔT and ΔT = 0 for isothermal process therefore ΔU = 0.

     

  • Question 3
    4 / -1

    For the reaction, 2Cl(g) ⟶ Cl2(g), the correct option is :

    Solution

    We know that, Cl2(g) ⟶ 2Cl(g) is endothermic reaction because it required energy to break bond.

    So reverse reaction, 2Cl(g) ⟶ Cl2(g) will be exothermic, ∆rH < 0 Also, two gaseous atom combine together to form 1 gaseous molecule.

     

  • Question 4
    4 / -1

    In which case change in entropy is negative?

    Solution

    If ∆ng < 0 then ∆S < 0

     

  • Question 5
    4 / -1

    Reversible expansion of an ideal gas under isothermal and adiabatic conditions are as shown in the figure.

    AB → Isothermal expansion AC → Adiabatic expansion

    Which of the following options is not correct?

    Solution

    In adiabatic expansion cooling effect will take place, TC will be less then TA.

    in adiabatic expansion q = 0

    ∆U = w

    Wpv < 0

    ∆U < 0

    nCvm ∆T < 0

    ∆T < 0

    TC − TA < 0

    $T_{C}

     

  • Question 6
    4 / -1

    For a given reaction, Δ H = 35.5 kJmol-1 and ΔS = 83.6JK-1mol (- 1) . The reaction is spontaneous at : (Assume that Δ H and Δ S do not vary with temperature)

    Solution

    Since the reaction is endothermic it will be spontaneous at T > 425K.

     

  • Question 7
    4 / -1

    For a sample of perfect gas when its pressure is changed isothermally from pi to pf, the entropy change is given by

    Solution

    For an ideal gas undergoing reversible expansion, when temperature changes from Ti to Tf and pressure changes from pi to pf,

     

  • Question 8
    4 / -1

    Consider the following liquid-vapour equilibrium. Liquid ⇌ Vapour Which of the following relations is correct?

    Solution

    This is Clausius-Clapeyron equation.

     

  • Question 9
    4 / -1

    For the reaction X2O4(l) → 2XO2(g) ΔU = 2.1 kcal, ΔS = 20 cal K−1 at 300K Hence, ΔG is

    Solution

    ΔH =ΔU + ΔngRT

    Given, ΔU = 2.1 kcal,Δng = 2

    R = 2 × 10−3 kcal, T = 300K

    ∴ ΔH = 2.1 + 2 × 2 × 10−3 × 300 = 3.3 kcal

    Again ΔG = ΔH − TΔS

    Given, ΔS = 20 × 10−3kcalK−1

    On putting the value of ΔH in the equation, we get

    ΔG = 3.3 − 300 × 20 × 10−3

    = 3.3 − 6 × 10× 10−3 = −2.7kcal

     

  • Question 10
    4 / -1

    A reaction having equal energies of activation for forward and reverse reactions has

    Solution

    ΔH = (Ea)− (Ea)b = 0

     

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