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Heat of combustion

  1. Apr 4, 2016 #1
    1. The problem statement, all variables and given/known data
    The combustion of hydrazine, N2H4, produces nitrogen gas and water vapor. The heat of combustion for this reaction is -618 kJ/mol. If 1.6 g of hydrazine are combusted in a bomb calorimeter at 298 K and with a heat capacity of 6.2 kJ/C, what will be the temperature of the bomb calorimeter after the reaction?

    Answer: 303 K

    2. Relevant equations
    Q=CΔT

    3. The attempt at a solution
    We're given ΔH, which is -618 kJ/mol. Thus 1.6 grams of hydrazine is 0.05 mol. Multiplying ΔH and mol gives me the heat evolved (Q) = -30.9 kJ.

    Setting up the equation, I get:
    Q=CΔT
    -30.9 kJ = (6.2 kJ/C)(Tf-25)
    Tf = 20 C

    Converting this to Kelvin, I get 293 K. What did I do wrong here? I think that I am confusing ΔH and Q?
     
  2. jcsd
  3. Apr 4, 2016 #2

    Ygggdrasil

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    Would you expect the combustion of a substance to increase or decrease the temperature of the calorimeter?
     
  4. Apr 4, 2016 #3
    The calorimeter and its contents is assumed to be adiabatic, so Q = 0. The change in internal energy for the combination of calorimeter and contents is thus also zero (constant volume). From the heat of combustion, do you know how to get the change in internal energy of the reaction at 25 C? What is the change in the number of moles for the reaction?

    Chet
     
  5. Apr 5, 2016 #4
    Increase since it is exothermic. My numbers are off.
     
  6. Apr 5, 2016 #5
    Not only are you numbers off. Your equation is off.

    Chet
     
  7. Apr 5, 2016 #6
    So we know that ΔH is -618 kJ/mol. Given that there is 1.6 g of hydrazine, this translate in to 0.05 mol. Thus, Q = 31 kJ. If one mole of hydrazine was combusted, the heat of combustion tells us that 618 kJ of energy is released.

    I understand that this is exothermic, thus the temperature must increase. I can do process of elimination for this problem but I would like to understand where my reasoning was flawed.

    I know that I have to use Q=CΔT for a bomb calorimeter.
     
  8. Apr 5, 2016 #7
    The ##\Delta H## referred to by the heat of combustion does not refer to the ##\Delta H## of the calorimeter. The heat of combustion is the amount of heat that must be removed from a combination of 1 mole of N2H2 and 1 mole of oxygen when they react completely at 1 atm to form 1 mole of N2 and 2 moles of water, in order to hold the temperature constant at 25 C. Even if we neglect the change in the number of moles, because the calorimeter is insulated, ##\Delta H## for the combination of reaction mixture and calorimater is equal to Q for the combination of reaction mixture and calorimeter, and both are equal to zero. So, for the combination of calorimeter and reaction mixture, ##\Delta H=0##. If the reaction mixture started at 25 C, and its final temperature were 25 C, its change in enthalpy would be -31 kJ. The overall change in enthalpy for the reaction mixture plus the calorimeter is then $$\Delta H=-31 + C (T-25) = 0$$
     
  9. Apr 5, 2016 #8
    So that's what I was missing: Qrxn=-Qcal
     
  10. Apr 5, 2016 #9
    Yes, if you want to say it that way. But I always reserve the use of the symbol Q exclusively for heat transferred in or out of a system through the boundary between the system and the surroundings. This is how the symbol Q is used in thermodynamics.

    Chet
     
  11. Apr 5, 2016 #10

    Ygggdrasil

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    Science Advisor

    In these types of equations, I always say to use your calculator to get the numbers, but use your brain to figure out the signs. There are so many examples where a negative sign flips when transferring numbers between different equations that it's always good to do a sanity check (e.g. thinking about whether the temperature should go up or down) to figure out if you need to carry a negative sign between equations.
     
  12. Apr 5, 2016 #11
    We should also mention that it isn't clear from the problem statement whether the heat of combustion represents the "higher heating value" or the "lower heating value." In the former case, the water is assumed to be liquid in the final state, and the number of moles of gas decreases, while, in the latter case, the water is assumed to be vapor and the number of moles of gas increases. These considerations are important in determining the internal energy change of the reaction, which may or may not include heat of condensation and which also must include the ##Delta (PV)##. For a constant volume system like this, it is the internal energy change that is zero, not the enthalpy change. Also, the problem statement does not indicate whether the sensible heat change of the gas mixture is expected to be included.

    Chet
     
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