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    Hey,

    Could anyone please help me with this question:

    A calorimeter was calibrated by burning 2.00 g of methanol (CH3OH) whose enthalpy of combustion is -715KJmol-1. The temperature of the calorimeter rose from 19.6°C to 52.4°C. The same calorimeter was used to measure the enthalpy of combustion of propan-2-ol. 1.50 g of propan-2-ol CH3CH(OH)CH3 raised the temperature from 19.8C to 56.2C. Calculate the heat capacity of the calorimeter and then the enthalpy of combustion of propan-2-ol.

    A step by step method and explanation would be much appreciated!

    Cheers
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    (Original post by Mooodle)
    Hey,

    Could anyone please help me with this question:

    A calorimeter was calibrated by burning 2.00 g of methanol (CH3OH) whose enthalpy of combustion is -715KJmol-1. The temperature of the calorimeter rose from 19.6°C to 52.4°C. The same calorimeter was used to measure the enthalpy of combustion of propan-2-ol. 1.50 g of propan-2-ol CH3CH(OH)CH3 raised the temperature from 19.8C to 56.2C. Calculate the heat capacity of the calorimeter and then the enthalpy of combustion of propan-2-ol.

    A step by step method and explanation would be much appreciated!

    Cheers
    1. work out the energy produced by burning 1 mol of methanol
    2. what percentage is this of the enthalpy of combustion
    3. apply the same percentage for the energy released by burning the propan-2-ol
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    (Original post by Mooodle)
    Hey,

    Could anyone please help me with this question:

    A calorimeter was calibrated by burning 2.00 g of methanol (CH3OH) whose enthalpy of combustion is -715KJmol-1. The temperature of the calorimeter rose from 19.6°C to 52.4°C. The same calorimeter was used to measure the enthalpy of combustion of propan-2-ol. 1.50 g of propan-2-ol CH3CH(OH)CH3 raised the temperature from 19.8C to 56.2C. Calculate the heat capacity of the calorimeter and then the enthalpy of combustion of propan-2-ol.

    A step by step method and explanation would be much appreciated!

    Cheers
    Where you given the mass/(density and volume) of liquid used inside the calorimeter?
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    (Original post by YouMadBro!)
    Where you given the mass/(density and volume) of liquid used inside the calorimeter?
    No... the information in the question above was all the information provided
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    (Original post by charco)
    1. work out the energy produced by burning 1 mol of methanol
    2. what percentage is this of the enthalpy of combustion
    3. apply the same percentage for the energy released by burning the propan-2-ol
    can you explain that method in simpler steps please?
    I got that, from the experiments, 1 mol of methanol would lead to -11440kJ mol-1 . Is this the right sort of line?
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    I think I got it!
    I wrote is down but thought it was dumb idea but now think it's the right answer

    For methanol,
    Enthalpy change = -(mass x heat capactiy x change in temperature) / mol
    Putting the values in for methanol
    -715 = -(mass x heat cap x 32.8) / (2/32)
    -44.6875...= -32mc
    1.363..=mc
    Since you aren't given the mass of the liquid, I think you can take it to mean that it's included into the heat capacity as you can assume that the settings of the calorimeter does not change.

    For propan-2-ol,
    just put in the value you got for (mass x heat cap) to work out enthaply change.
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    (Original post by YouMadBro!)
    I think I got it!
    I wrote is down but thought it was dumb idea but now think it's the right answer

    For methanol,
    Enthalpy change = -(mass x heat capactiy x change in temperature) / mol
    Putting the values in for methanol
    -715 = -(mass x heat cap x 32.8) / (2/32)
    -44.6875...= -32mc
    1.363..=mc
    Since you aren't given the mass of the liquid, I think you can take it to mean that it's included into the heat capacity as you can assume that the settings of the calorimeter does not change.

    For propan-2-ol,
    just put in the value you got for (mass x heat cap) to work out enthaply change.
    Bless your soul sweet man.
    May fortune smile upon you for the rest of your days.
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    I made an account on this website just to personally thank you for your genius. I have no idea how I didn't think to remove M from q=MS^T but thank you!
    (Original post by YouMadBro!)
    I think I got it!
    I wrote is down but thought it was dumb idea but now think it's the right answer

    For methanol,
    Enthalpy change = -(mass x heat capactiy x change in temperature) / mol
    Putting the values in for methanol
    -715 = -(mass x heat cap x 32.8) / (2/32)
    -44.6875...= -32mc
    1.363..=mc
    Since you aren't given the mass of the liquid, I think you can take it to mean that it's included into the heat capacity as you can assume that the settings of the calorimeter does not change.

    For propan-2-ol,
    just put in the value you got for (mass x heat cap) to work out enthaply change.
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    (Original post by georgewood0)
    I made an account on this website just to personally thank you for your genius. I have no idea how I didn't think to remove M from q=MS^T but thank you!
    Your welcome. I wish you the best of luck in the hell others call A-Level Chemistry. xD
 
 
 
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