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Thermodynamics Test - 59

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Thermodynamics Test - 59
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  • Question 1
    1 / -0
    What are the most favourable conditions for the reaction;
    SO2(g)+12O2(g)SO3(g);ΔHo=ve{ SO }_{ 2 }(g)+\cfrac { 1 }{ 2 } { O }_{ 2 }(g)\leftrightharpoons { SO }_{ 3 }(g);\Delta { H }^{ o }=-ve to occur?
    Solution
    Negative value of enthalpy change suggests exothermic reaction. Such reactions are favored at low temperature as heat released during the reaction raised the temperature and nullifies the effect of low temperature.
    Δn= number of moles of gaseous  products   number of moles of gaseous reactants       \displaystyle \Delta n = \text { number of moles of gaseous  products } -  \text { number of moles of gaseous reactants  }    
    Δn=1(1+1  2) \displaystyle \Delta n = 1 - (1+ \dfrac {1  }{  2 } )
    Δn=1  2 \displaystyle \Delta n =-\dfrac {1  }{  2 }
    SinceΔn <0 \displaystyle \Delta n  < 0, the reaction is favored at high pressure. Increasing the pressure will shift the reaction towards forward direction (one with less number of moles of gaseous species). This will nullify the effect of increase in pressure.
  • Question 2
    1 / -0
    Given, 
    C(s)+O2(g)CO2(g);ΔH=395 kJC(s)+{O}_{2}(g)\rightarrow {CO}_{2}(g); \Delta H=-395\ kJ
    S(s)+O2(g)SO2(g);ΔH=295 kJS(s)+{O}_{2}(g)\rightarrow {SO}_{2}(g);\Delta H=-295\ kJ
    CS2(l)+3O2(g)CO2(g)+2SO2(g);ΔH=1110 kJC{S}_{2}(l)+3{O}_{2}(g)\rightarrow {CO}_{2}(g)+2{SO}_{2}(g);\Delta H=-1110\ kJ

    The heat of formation of CS2(l)C{S}_{2}(l) is:
    Solution
    Since, the formation reaction of CS2CS_2 is,

    C(s)+2S(s)CS2(l)C(s) + 2S(s) \rightarrow CS_2(l)

    Since, the formation reaction of CS2CS_2 can also be derived using the given reaction as,

    1streaction+2×2nd reaction3rd reaction1^{st} \text{reaction} +2\times 2^{nd} \text{ reaction} - 3^{rd} \text{ reaction}

    Hence, the enthalpy of the reaction,
     
    [1streaction+2×2nd reaction3rd reaction][1^{st} \text{reaction} +2\times 2^{nd} \text{ reaction} - 3^{rd} \text{ reaction}] =ΔH1streaction+2×ΔH2nd reactionΔH3rd reaction=\Delta H_{1^{st} \text{reaction}} +2\times \Delta H_{2^{nd} \text{ reaction}} -\Delta H_{3^{rd} \text{ reaction}} =heat of formation of CS2(l)=\text{heat of formation of CS}_2(l)

    Hence, ΔHf(CS2)=395+2×(295)(1110)=125kJ \Delta H_{f(CS_2)} = -395 + 2\times (-295) - (-1110) = 125kJ

    Hence, the answer is option DD.
  • Question 3
    1 / -0
    If ΔHf(H2O)\Delta H_{f}(H_{2}O) is 285.20 kJ mol1-285.20\ kJ\ mol^{-1}, then ΔHf(OH)\Delta H_{f}^{\circ}(OH^{-}) is:
    Solution
    H++OHH2O,ΔH=57.3 kJ mol1H^{+} + OH^{-}\rightarrow H_{2}O, \Delta H = -57.3\ kJ\ mol^{-1} (standard value)

    ΔH=ΔHf(H2O)ΔHf(H+)ΔHf(OH)\Delta H = \Delta H_{f}^{\circ} (H_{2}O) - \Delta H_{f}^{\circ} (H^{+}) - \Delta H_{f}^{\circ}(OH^{-})

    57.3=285.20ΔHf(OH)-57.3 = -285.2 - 0 - \Delta H_{f}^{\circ}(OH^{-})

    ΔHf(OH1)=227.9 kJ mol1\Delta H_{f}^{\circ} (OH^{-1}) = -227.9\ kJ\ mol^{-1}.
  • Question 4
    1 / -0
    The enthalpy of combustion for the H2H_{2}, cyclohexene and cyclohexane are 241,3800-241, -3800 and 3920kJ mol1-3920kJ\ mol^{-1}, respectively. Heat of hydrogenation of cyclohexene is:
    Solution
    Different combustion reactions are shown below.
    C6H12+9O26CO2+6H2O\displaystyle C_6H_{12}+9O_2 \rightarrow 6CO_2+6H_2O   

    ΔH1=3920\displaystyle \Delta H_1 = -3920 kJ/mol   ......(1)

    C6H10+8.5O26CO2+5H2O\displaystyle C_6H_{10}+8.5O_2 \rightarrow 6CO_2+5H_2O   

    ΔH2=3800\displaystyle \Delta H_2 = -3800 kJ/mol   ......(2)

    H2+O2H2O\displaystyle H_{2}+O_2 \rightarrow H_2O   ΔH3=241\displaystyle \Delta H_3 = -241 kJ/mol   ......(3)

    The balanced chemical equation for the hydrogenation of cyclohexene is

    Cyclohexene +H2\displaystyle + H_2 \rightarrow Cyclohexane   .....(4)

    Add reaction (3) to reaction (2). Then substract reaction (1) to obtain reaction (4).

     ΔH4=ΔH2ΔH1+ΔH3\displaystyle  \Delta H_{4} = \Delta H_2- \Delta H_1+ \Delta H_3
     ΔH4=3800(3920)+241\displaystyle  \Delta H_4=-3800-(-3920)+-241
     ΔH4=121\displaystyle  \Delta H_4=-121 kJ/mol

    Hence, the heat of hydrogenation of cyclohexene is  ΔH=121\displaystyle  \Delta H=-121 kJ/mol
  • Question 5
    1 / -0
    Standard heats of formation for CCl4,H2O,CO2C{Cl}_{4},{H}_{2}O,{CO}_{2} and HClHCl at 298K298K are 25.5,57.8,94.1-25.5,-57.8,-94.1 and 22.1kJ/mol-22.1kJ/mol respectively.
    For the reaction, what will be ΔH\Delta H?
    CCl4+2H2OCO2+4HClC{Cl}_{4}+2{H}_{2}O\rightarrow {CO}_{2}+4HCl
    Solution
    For the given reaction,
    CCl4+2H2OCO2+4HClCCl_4 +2H_2O \rightarrow CO_2 + 4HCl
    ΔH=ΔHf(CO2)+\Delta H = \Delta H_{f(CO_2)} + 4×4\times ΔHf(HCl)\Delta H_{f(HCl)} - ΔHf(CCl4)\Delta H_{f(CCl_4)} - 2×2\times ΔHf(H2O)\Delta H_{f(H_2O)}
    => ΔH=94.1+4×(22.1)(25.5)2×(57.8)=41.4KJ\Delta H = -94.1 + 4\times (-22.1) - (-25.5) - 2\times (-57.8) = -41.4 KJ
    Hence, answer is option DD.
  • Question 6
    1 / -0
    Water is brought to boil under the pressure of 1.0 atm1.0\ atm. When an electric current of 0.50 A0.50\ A from a 12 V12\ V supply is passed for 300 s300\ s through resistance in thermal contact with it is found that 0.798 g0.798\ g of water is vapourised. Calculate the molar internal energy change at boiling point (373.15 K)(373.15\ K).
    Solution
    ΔH=\Delta H= work done=i×V×t=i\times V\times t
             =0.5×12×300=1.8 kJ=0.5\times 12\times 300=1.8\ kJ

    Molar enthalpy of vaporisation
    ΔHm=ΔHmolesofH2O =ΔHnH2O\Delta { H }_{ m }=\cfrac { \Delta H }{ moles\quad of\quad { H }_{ 2 }O\quad  } =\cfrac { \Delta H }{ { n }_{ { H }_{ 2 }O } }

               =1.8kJ0.79818 =40.6 kJmol1=\cfrac { 1.8kJ }{ \cfrac { 0.798 }{ 18 }  } =40.6\ kJ{ mol }^{ -1 }

    ΔHm=ΔEm+pΔV\Delta { H }_{ m }=\Delta { E }_{ m }+p\Delta V

    ΔHm=ΔEm+ΔngRT\Delta { H }_{ m }=\Delta { E }_{ m }+\Delta { n }_{ g }RT\quad

    ΔHm=ΔEm+RT\Delta { H }_{ m }=\Delta { E }_{ m }+RT\quad

    [Δng=1forH2O(l)H2O(g)\therefore \Delta { n }_{ g }=1\quad for\quad { H }_{ 2 }O(l)\rightleftharpoons { H }_{ 2 }O(g)]

    \therefore Molar internal energy change,

    ΔEm=ΔHmRT\Delta { E }_{ m }=\Delta { H }_{ m }-RT

    ΔEm=40.68.314×103×373.15=37.5 kJmol1\Delta { E }_{ m }=40.6-8.314\times { 10 }^{ -3 }\times 373.15=37.5\ kJ { mol }^{ -1 }
  • Question 7
    1 / -0
    Given: S+32O2SO3+2xS+\dfrac { 3 }{ 2 } { O }_{ 2 }\longrightarrow S{ O }_{ 3 }+2x kcalkcal
               SO2+12O2SO3+yS{ O }_{ 2 }+\dfrac { 1 }{ 2 } { O }_{ 2 }\longrightarrow S{ O }_{ 3 }+y kcalkcal

    With the help of the above reactions, find out the heat of formation of SO2S{ O }_{ 2 }.
    Solution
    S+32O2SO3+2xkcalS+\dfrac { 3 }{ 2 } { O }_{ 2 }\longrightarrow S{ O }_{ 3 }+2x \: kcal.....(1)

    SO2+12O2SO3+ykcalS{ O }_{ 2 }+\dfrac { 1 }{ 2 } { O }_{ 2 }\longrightarrow S{ O }_{ 3 }+y \: kcal......(2)

    The heat of formation of SO2S{ O }_{ 2 } can be represented as $$S+\  { O }_{ 2

    }\longrightarrow S{ O }_{2 } ; \: \ \ \ \ \ \Delta H_f(SO_2)$$......(3)

    Subtract equation (2) from equation (1) to obtain equation (3), we get-

     ΔHf(SO2)=(2xy) \displaystyle  \Delta H_f(SO_2)=\left( 2x-y \right) kcal

    Hence, option A is correct.
  • Question 8
    1 / -0
    A particular reaction at 27oC{27}^{o}C for which ΔH>0\Delta H > 0 and ΔS>0\Delta S> 0 is found to be non-spontaneous. The reaction may proceed spontaneously if:
    Solution
    Gibb's energy equation: ΔG=ΔHTΔS\Delta G=\Delta H-T\Delta S
    For a spontaneous reaction, ΔG\Delta G should be less than zero. 

    Therefore, if ΔH>0\Delta H>0 and ΔS>0\Delta S>0, the TΔST\Delta S term should be higher than ΔH\Delta H for a spontaneous reaction. So, the temperature should be increased to make the reaction spontaneous.
  • Question 9
    1 / -0
    Although the dissolution of ammonium chloride in water is an endothermic reaction, even then it is spontaneous because:
    Solution
    Gibb's free energy equation: ΔG=ΔHTΔS\Delta G=\Delta H-T\Delta S

    For a spontaneous reaction, ΔG<0\Delta G<0
    For endothermic reaction, ΔH<0\Delta H<0

    For dissolution of ammonium chloride in water, ΔS>0\Delta S>0. The magnitude of TΔST\Delta S should be greater than that of ΔH\Delta H so that ΔG<0\Delta G<0.

    Therefore, the dissociation of ammonium chloride in water is an endothermic reaction but is still spontaneous because ΔH\Delta H is +ve+veΔS\Delta S is +ve+ve and ΔH<TΔS\Delta H<T\Delta S.
  • Question 10
    1 / -0
    The enthalpy of hydrogenation of benzene is 49.8 kcal/mol-49.8\ kcal/mol while its resonance energy is 36.0 kcal/mol36.0\ kcal/mol. The enthalpy of formation of benzene is:
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