For the reaction P40 10(s) + 6 H20(1) 4 H3PO4(aq) AG° = -448.9 kJ and AS° = -16.3 J/K at 261 K and 1 atm. This reaction is (reactant, product) favored under standard conditions 261 K. The standard enthalpy change for the reaction of 2.15 moles of P4010(s) at this temperature would be kJ.

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Chapter6: Thermochemisty
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Problem 6.132QP
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For the reaction
P4010(s) + 6 H20(1)
4 H3PO4(aq)
AG° = -448.9 kJ and AS° = -16.3 J/K at 261 K and 1 atm.
This reaction is (reactant, product)
favored under standard conditions at 261 K.
The standard enthalpy change for the reaction of 2.15 moles of P4010(s) at this temperature would be
kJ.
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Transcribed Image Text:For the reaction P4010(s) + 6 H20(1) 4 H3PO4(aq) AG° = -448.9 kJ and AS° = -16.3 J/K at 261 K and 1 atm. This reaction is (reactant, product) favored under standard conditions at 261 K. The standard enthalpy change for the reaction of 2.15 moles of P4010(s) at this temperature would be kJ. Submit Answer Try Another Version 1 item attempt remaining rch a
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