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24 September, 03:38

When 1.50 g of ba (s) is added to 100.00 g of water in a container open to the atmosphere, the reaction shown below occurs and the temperature of the resulting solution rises from 22.00°c to 33.10°c. if the specific heat of the solution is 4.18 j / (g ∙ °c), calculate δh for the reaction, as written. ba (s) + 2h2o (l) →ba (oh) 2 (aq) + h2 (g) δh=?

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Answers (2)
  1. 24 September, 04:57
    0
    Given:

    Mass of Ba = 1.50 g

    Mass of H2O = 100.0 g

    Initial temp T1 = 22 C

    Final Temp T2 = 33.1 C

    specific heat c = 4.18 J/g c

    To determine:

    The reaction enthalpy

    Explanation:

    The heat released during the reaction is:

    q = - mc (T2-T1) = - (100+1.5) g * 4.18 J/g C * (33.1-22) C = - 4709.4 J

    # moles of Ba = Mass of Ba/Atomic mass of Ba = 1.5 g/137 g. mol-1 = 0.0109 moles

    ΔH = q/mole = - 4709.4 J/0.0109 moles = - 432 kJ/mol

    Ans : The enthalpy change for the reaction is - 432 kJ/mol
  2. 24 September, 05:47
    0
    The enthalpy of reaction = - 428.127 kJ

    Explanation:

    Enthalpy of a reaction is the amount of heat change during a reaction of one mole of a substance.

    The heat released in the reaction is being absorbed by the solution.

    The heat absorbed by solution = mass of solution X specific heat X change in temperature

    Heat absorbed = (100+1.5) X 4.18 X (33.10-22) = 4709.397 Joules

    This heat being released when 1.50g of Barium is added to water

    The moles of Barium added = mass / atomic mass = 1.5 / 137.33 = 0.011 moles

    the heat released on adding 0.011 moles of barium = 4709.397 Joules

    So heat released on adding 1 mole of barium = 428127 Joules = 428.127 kJ

    The enthalpy of reaction = - 428.127 kJ
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