When 0.1523 g of liquid pentane (CH) combusts in a bomb calorimeter, the temperature rises from 23.7C to 29.8 C. What is U for the reaction

Question

When 0.1523 g of liquid pentane (CH) combusts in a bomb calorimeter, the temperature rises from 23.7C to 29.8 C. What is U for the reaction in kJ/mol pentane? The heat capacity of the bomb calorimeter is 5.23 kJ/C.

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Euphemia 3 years 2021-08-24T04:38:32+00:00 1 Answers 48 views 0

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  1. minhkhue
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    2021-08-24T04:39:43+00:00
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    Answer:

    U for the reaction is 15,048.58 kJ/mol of pentane.

    Explanation:

    Quantity of heat required = heat capacity of bomb calorimeter × temperature rise = 5.23 kJ/C × (29.8 C – 23.7 C) = 5.23 kJ/C × 6.1 C = 31.903 kJ

    Moles of pentane = mass/MW

    Mass = 0.1523 g

    MW of pentane (C5H12) = 72 g/mol

    Moles of pentane = 0.1523/72 = 0.00212 mol

    U for the reaction = quantity of heat required ÷ moles of pentane = 31.903 kJ ÷ 0.00212 mol = 15,048.58 kJ/mol

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