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champagnecaribou797Lv1

5 Apr 2020

(a) Why are tables of standard enthalpies of formation so useful? (b) What is the value of the standard enthalpy of formation of an element in its most stable form? (c) Write the chemical equation for the reaction whose enthalpy change is the standard enthalpy of formation of sucrose (table sugar), C₁₂H₂₂O₁₁(s), ∆H_f°[C₁₂H₂₂O₁₁].

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Bunny GreenfelderLv2

20 May 2020

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COULD SOMEONE PLEASE HELP ME ANSWER THESE QUESTIONS AND IF POSSIBLE EXPLAIN?

1. Standard enthalpies of formation are usually computed from calorimetric data referring to reactions which are easy to carry out in the laboratory. For many compounds, the enthalpy of the reaction with O₂ (g) (combustion) is commonly used, together with the standard enthalpies of formation of the combustions products, to compute ?H_f⁰. For example, sucrose, C₁₂H₂₂O₁₁, common table sugar, is found to have a standard enthalpy of combustion of -5640.9 kJ mole^-1. The standard enthalpies of formation of CO₂ (g) and H₂O (l) are, respectively, -393.51 and -285.83 kJ mole^-1. Compute ?H_f⁰ for sucrose.

2. A student determined the enthalpy of solution of NaOH by dissolving 4.00 g of NaOH (s) in 180 ml of pure water (density 0.999 g ml^-1) at an initial temperature of 19.5

The enthalpy changes referred to by equations:
1/2 N2(g)+3/2 H2(g)+ x H2O(l)>>NH3 (aq)

1/2 H2(g)+ 1/2 Cl2 (g)+y H2O(l)>> HCl (aq)

and given in table 3 are not strictly standard enthalpies of formation since the products are not in standard states(though the reactants are). the measured enthalpies of neutralization and dissolution are not standard enthalpies either. yet there is an equation states that the standard enthalpy of formation is calculated by summing for non-standard enthalpy changes. explain why this is so?

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