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Part II: Estimating Heat of Reaction from ?H<sup>o</sup> Find the value of ?H<sup>o</sup> for the reaction H<sub>2</sub>O<sub>2</sub> (l) ? H<sub>2</sub>O (l) + 1/2 O<sub>2</sub> (g) from these values of ?H<sup>o</sup><sub>f</sub>: H<sub>2</sub>O (l) = -286 kJ/mol H<sub>2</sub>O<sub>2</sub> = -188 kJ/mol (show calculation) ?H<sup>o</sup> _______ kJ/mol

          Part II: Estimating Heat of Reaction from ?H<sup>o</sup>
Find the value of ?H<sup>o</sup> for the reaction H<sub>2</sub>O<sub>2</sub> (l) ? H<sub>2</sub>O (l) + 1/2 O<sub>2</sub> (g)
from these values of ?H<sup>o</sup><sub>f</sub>: H<sub>2</sub>O (l) = -286 kJ/mol  H<sub>2</sub>O<sub>2</sub> = -188 kJ/mol
(show calculation)
?H<sup>o</sup> _______ kJ/mol
        
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Part II: Estimating Heat of Reaction from ?H<sup>o</sup>
Find the value of ?H<sup>o</sup> for the reaction H<sub>2</sub>O<sub>2</sub> (l) ? H<sub>2</sub>O (l) + 1/2 O<sub>2</sub> (g)
from these values of ?H<sup>o</sup><sub>f</sub>: H<sub>2</sub>O (l) = -286 kJ/mol  H<sub>2</sub>O<sub>2</sub> = -188 kJ/mol
(show calculation)
?H<sup>o</sup>  kJ/mol

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Chemistry: Structure and Properties
Chemistry: Structure and Properties
Nivaldo Tro 2nd Edition
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Part II: Estimating Heat of Reaction from ΔHr Find the value of ΔH for the reaction H2O(l) → H2O(g) from these values of ΔHf H2O(l) = -286 kJ/mol and ΔHf H2O(g) = -188 kJ/mol (show calculation) ΔH = ΔHf H2O(g) - ΔHf H2O(l) ΔH = -188 kJ/mol - (-286 kJ/mol) ΔH = -188 kJ/mol + 286 kJ/mol ΔH = 98 kJ/mol
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00:01 Hello students in this question for the given values of delta h not we have to calculate the enthalpy of the reaction that is delta h0 for the reaction at temperature t equals to 298 kelvin okay so delta h not for the reaction can be written for the particular so 2 multiplied by delta h not h1 plus 2 multiplied by delta h20 minus delta h3 0.
00:31 Okay, so we have the given value.
00:33 So 2 modipater by delta h1, not, which is minus of 537 kilojoule, plus 2 made delta h2, which is minus 680 kilojoule, minus of delta h3, which is 52 kilojoule.
00:48 Okay, so we can solve them...
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