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Thermodynamics Test - 76

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Thermodynamics Test - 76
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  • Question 1
    1 / -0
    For the reaction  : 

    $$3N_{2} + H_{2} \rightarrow 2NH_{3}$$,   

    $$\triangle H = - 24 kcal$$ at $$427^{0}C$$ and $$200 atm$$

    Calculate magnitude of internal enerrgy change in $$Kcal$$ ($$\triangle U$$), if 168 g $$N_{2}$$ gas and 30 g $$H_{2}$$ gas are allowed to react completely (100% reaction yield) to form $$NH_{3}$$ gas at $$427^{0}C$$ and 200 atm. 
  • Question 2
    1 / -0
    Which of the following are spontaneous?
  • Question 3
    1 / -0
    One mole of an ideal monatomic gas undergoes a process described by the equation $$PV^{3}$$ = constant. The heat capacity of the gas during this process is:
  • Question 4
    1 / -0
    A reaction A(g) +B(g) $$\rightarrow $$ C(g) +D(g). $$\Delta H$$= (-) ve is found to have positive entropy change. the reaction will be: 
    Solution

  • Question 5
    1 / -0
    The correct thermodynamic conditions for the spontaneous reaction at all temperatures is ______________________.
    Solution

  • Question 6
    1 / -0
    7.5 KJ heat is added to a closed system and its internal energy decreases by 12 KJ. So,______ of work is done ____ the system ,for a new process , if the work is zero, then____of heat is ______ for the same changes in state of system
  • Question 7
    1 / -0
    An ideal monoatomic gas is taken through a process in which $$d Q = 2 d U$$ . The molar heat capacity of the gas for the process is
    Solution

  • Question 8
    1 / -0
    Correct desending order of deprotonation in the following compound:

  • Question 9
    1 / -0
    $$100g$$ of water is heated from $${30}^{o}C$$ to $${50}^{o}C$$. Ignoring the slight expansion of water, the change in its internal energy is (Specific heat of water is $$4200J$$ $${kg}^{-1}$$ $${K}^{-1}$$)
    Solution

  • Question 10
    1 / -0
    If $$\Delta G_{298}$$ for reaction:
    $$2H_{2 (9, 1 atm)} + O_{2(g)(1atm)} \to 2H_2O(g)(1 atm)$$
    is $$-240KJ$$, what will be the $$\Delta G_{298}$$ for reaction
    $$H_2O_{(g, 0.2atm)} \to H_{2(g, 4atm)} + \dfrac{1}{2}O_{2(g, 0.9 atm)}$$
    Solution
    For one mole $$H_2O$$ formation at  1atm
    $$= \dfrac{-240}{2} = -120 kJ$$
    For decomposition $$=+120KJ$$
    $$Kep = P \left[\dfrac{\sqrt{O_2} \times H_2}{420}\right] = \dfrac{\sqrt{0.25}\times 4}{0.2}$$
    $$= 0.5\times 20$$
    $$= 10$$

    $$\Delta G = \Delta G^o + RT\ ln K$$
    $$ = 120000+ 8.314\times 298 \times ln 10$$
    $$= 125705 J$$
    $$= 125.7 KJ$$ 
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