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C1210 lecture 15 - Suppose we know the easy-to-conduct...

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Suppose we know the easy-to-conduct reaction: C( s ) + O 2 ( g ) CO 2 ( g ) Δ H 0 = -393.5 kJ Because of conservation of energy we also know : CO 2 ( g ) C( s ) + O 2 ( g ) Δ H 0 = +393.5 kJ Even though we can’t run this (very energy uphill) reaction!!!!
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Let C and O 2 react: We’ll measure C( s ) + O 2 ( g ) CO 2 ( g ) Δ H 0 = -393.5 kJ Now conduct two, separate reactions and measure the Δ H 0 ’s: C( s ) + ½O 2 ( g ) CO( g ) Δ H 0 = -110.5 kJ CO( g ) + ½O 2 ( g ) CO 2 ( g ) Δ H 0 = -283.0 kJ Add ‘em up: C( s ) + ½O 2 ( g ) + CO( g ) + ½O 2 ( g ) CO( g ) + CO 2 ( g ) Δ H 0 = -110.5 - 283.0 kJ C(s) + O 2 (g) CO 2 ( g ) Δ H 0 = -393.5 kJ H is a state function, independent of path. • We can combine thermochemical equations to predict thermochemistry of reactions we haven’t observed!!!!!!!!!!!! A very big deal.
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