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NEET Chemistry Thermodynamics / रासायनिक उष्मागतिकी Question Bank Self Evaluation Test - Thermodynamics

  • question_answer
    Heat of neutralization of a strong acid HA and a weaker acid HB with KOH are \[-13.7\] and\[-12.7\text{ }k\text{ }cal\text{ }mo{{l}^{-1}}\]. When 1 mole of KOH was added to a mixture containing 1 mole each of HA and HB, the heat change was \[-13.5\] kcal. In what ratio is the base distributed between HA and HB.             

    A) 3 : 1                 

    B) 1 : 3 

    C) 4 : 1                 

    D) 1 : 4

    Correct Answer: C

    Solution :

    Let x mole of KOH be neutralized by the strong acid HA. Then, moles neutralized by \[HB=1-x\] Hence, \[-13.7\times x+\left( -12.7 \right)\times \left( 1-x \right)=-13.5\] \[\Rightarrow x=0.8;\,\,\,\,\,\,\frac{x}{1-x}=\frac{0.8}{0.2}=\frac{4}{1}4:1\]


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