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Calculate the ΔH of the following reaction: CO2(g) + H2(g)O -> H2CO3(g) if the standard values of ΔHf are as follows: CO2(g) : -393.509 KJ/mol, H2O(g): – 241.83 KJ/mol, and H2CO3(g) : – 275.2 KJ/mol.

Calculate the ΔH of the following reaction: CO2(g) + H2(g)O -> H2CO3(g) if the standard values of ΔHf are as follows: CO2(g) : -393.509 KJ/mol, H2O(g): – 241.83 KJ/mol, and H2CO3(g) : – 275.2 KJ/mol. Correct Answer +360.139 KJ

ΔH° = ƩΔvpΔH°f (products) – ƩΔvrΔH°f (reactants) so this means that you add up the sum of the ΔH’s of the products and subtract away the ΔH of the products: (- 275.2 KJ) – (-393.509 KJ + -241.83KJ) = (-275.2) – (-635.339) = + 360.139 KJ.

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