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How much heat is released at constant pressure if a 10.0 L tank containing 87.0 atm of hydrogen sulfide gas condenses at its boiling point of -60.0 oC? The enthalpy of vaporization of hydrogen sulfide is 18.7 kJ/mol at -60.0 oC. (R = 0.0821 L • atm/(K • mol))

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Answer:

929.4 kJ of heat is released.

Explanation:

The enthalpy of vaporization of hydrogen sulfide is 18.7 kJ/mol at -60.0 °C.

The data explains that 1 mol of H₂S uses 18.7 kJ of energy in vaporization.

By the Ideal Gas Law, we can know, how many moles do we have, of gas.

P. V = n . R . T

87 atm . 10L = n . 0.0821 L.atm/K.mol . 213 K

(87 atm . 10L) / (0.0821 K.mol/L.atm . 213 K) = 49.7 moles

1 mol ____ 18.7 kJ

49.7 moles _____ (49.7 . 18.7) = 929.4 kJ

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