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Chemistry

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The combustion of benzoic acid releases 26.38kJ/g and is often used to calibrate a bomb calorimete. The combustion of 1.045g of benzoic acid caused the temperature of the calorimeter to rise by 5.985 degreeC.Using the same calorimeter, a sample of 0.876g of C8H18 was burned. the temp increased by 8.518 degreeC. what is the molar enthalpy of octane?

I don't know how to go about doing this question. What formula should I used I listed all the given:

Benzoic acid
deltaH= 26.318kJ/g
m= 1.045g
delta T= 5.985= 278.98K

C8H18
m= 0.876
mm= 114.224
delta T= 8.15= 281.52K
delta H= ?

  • Chemistry -

    First you need to address the data. I see 26.318 kJ/g for delta H in the table but 26.38 in the question. Be sure to take care of that first. To calculate the calorimeter constant, that is
    mass x specific heat x delta T.

    q = delta H, 26,380 J (check that number) = 1.045 x Cp x 5.985. Solve for Cp.

    For the octanol,
    [mass octanol x specific heat octanol x delta T] + [Cp x delta T] = 0
    Solve for specific heat octanol which will be in units of J/g*C.
    Then change that to kJ/mol.
    Check my thinking. Check my work.

  • Chemistry -

    Okie i get Cp= 42178.8J/g delta C

    and for the octanol
    im using q= m*Cs*T
    q= 0.876*Cs*8.158
    q= 7.1395g degreeC * Cs

    In your work i don't understand thisequation:
    [mass octanol x specific heat octanol x delta T] + [Cp x delta T] = 0

  • Chemistry -

    Okie I get 36913.6J/g* degree C but how do I change that into kJ/mol?

  • Chemistry -

    36913.6 J/g x (grams octanol/mol) = ??
    In other words, look up the molar mass of octanol.

  • Chemistry -

    36913.6 J/g * 114g/mol= -4208150.4J/mol and to convert that to kJ= -4208150400 kJ/mol.

    the correct answer is -5116 kJ/mol

  • Chemistry -

    this is so confusing =(

  • Chemistry -

    You need to clarify.
    Is benzoic 26.38 or 26.318?
    Is the temperature rise 8.158, 8.518, or 8.15?

  • Chemistry -

    Benzoic is 26.38 and the temp rises 8.518

  • Chemistry -

    I did is again:

    Benzoic acid: q= m*Cs*T
    Cs = 4.217kJ/g°C
    q1 + q2 = 0
    [m*Cs*T] + [26.38kJ] = 0
    [.876g*Cs*8.518°C]+ [26.38kJ] = 0
    7.462g°C * Cs = -26.38kJ
    Cs= -3.535kJ/g°C

    to get the molar enthalpy:
    -3.535kJ/g°C * 114g/mol= 403.03kJ/mol* °C

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