Question

In order to determine the molar heat of neutralization of potassium hydroxide, 200cm³ of 1M Hydrochloric acid and 200cm³ Of IM Potassium hydroxide both at the same initial temperature were mixed and stirred continuously with a thermometer. The temperature of the resulting solution was recorded after every 30 seconds until the highest temperature of the solution was attained. Thereafter, the temperature of the solution was recorded for a further two minutes.

1.  
   1. Why was it necessary to stir the mixture of the two solutions?   
   2. Define molar heat of neutralization   
   3. Write an ionic equation for the reaction which took place    
2. If the initial temperature for both solution was 24.5°C and the highest temperature attained by the mixture was 30.9°C.
   Calculate the:
   1. Heat change for the reaction (specific heat capacity of the solution = 4.2Jg⁻¹K⁻¹and the density of the solution = 1.0g/cm³)    
   2. Molar heat at neutralization of potassium hydroxide.     
3. Explain how the value of the molar heat of neutralization obtained in the experiment would compare with the one that would be obtained if the experiment was repeated using 200cm³ of 1 Mammonium hydroxide instead of IM potassium hydroxide.  
4. Use the following information to answer the questions that follow:
   C(s) + O₂(g) →  CO₂(g)                               ∆H = -393KJ/mol
   H₂(g) + ½O₂(g) → H20(g)                           ∆H = -296KJ/mol
   C₄H₁₀+ ¹³/₂O₂(g) → 4CO₂(g)+ 5H₂0       ∆H =-2877K]/moi
   Calculate the heat of formation of butane.   

Answer 1

1.  
   1. To ensure heat is evenly distributed in the solution
   2. Enthalpy change when 1mole of a OH⁻ are neutralized by 1mole of H⁺ to form 1 mole of water
   3. OH⁻(aq)+ H⁺(aq)→H₂O(l)
2.  
   1. ∆H =400 × 4.2 × 6.4
           =10752J//10.75%
   2. Moles of KOH =200 × 1
                                 1000
                             =0.2
       0.2→10.752
      1mole→?
      10.752 × 1
            0.2
      =−53.76KJ/mole
3. Ammonium hydroxide is a weak base, it's partially ionized some of the heat is used up to fully ionize. Heat of KOH is higher than ammonium hydroxide
4. 
    
   ∆H=4x − 393 + 5x −296−−2877
       =−1572+−1480−−2877
   =−175KJ/mole