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JEE Main & Advanced Sample Paper JEE Main Sample Paper-1

  • question_answer
    The standard heat of combustion of carbon(s), sulphur (s) and carbon disulphide \[(l)\] are-393.3, -293.72 and - 1108.76 kJ/mol respectively. The standard heat of formation of carbon disulphide (0 is

    A) -128.02 kJ                           

    B) +128.02kJ

    C) -218.42 kJ                           

    D)  + 218.42 kJ

    Correct Answer: B

    Solution :

     Given, I.\[C(s)+{{O}_{2}}\to C{{O}_{2}}\Delta H=-393.3\,\text{kJ/mol}\] II.\[S(s)+{{O}_{2}}\to S{{O}_{2}}\Delta H=-293.72\,\text{kJ/mol}\] III.\[C{{S}_{2}}(l)+3{{O}_{2}}(g)\to C{{O}_{2}}(g)+2S{{O}_{2}}(g)\] \[\Delta H=-110.8.76\,\text{kJ/mol}\] On putting various enthalpy of formation in equation III\[\Delta H=\Delta H\] (products) \[-\Delta H\] (reactants)-1108.76 = [- 393.3 + 2(- 293.72)] \[-[\Delta {{H}_{f}}(C{{S}_{2}})+3\times 0]\] \[-1108.76=-393.3-2\times 293.72-\Delta {{H}_{f}}(C{{S}_{2}})\] \[\Delta {{H}_{f}}(C{{S}_{2}})=128.02\text{kJ}\]


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