Explain the standard enthalpy change for the reaction

Reference no: EM13335596

2CO(g) + O2(g) 2CO2(g) Go = -506.5 kJ and So = -173.1 J/K at 344 K and 1 atm. This reaction is product favored under standard conditions at 344 K. The standard enthalpy change for the reaction of 1.85 moles of CO(g) at this temperature would be -524 kJ. Ho = -506.5 + 344 * -173.1 divide it all by 1000= -566.0kj i dont understand how they got -566.0kj, when i put it in my calculator i keep getting 59.0.. can you tell me what im doing wrong.

Reference no: EM13335596

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