Lecture 16sf

# Lecture 16sf - Combining Reactions to Calculate H N2 O2 2NO...

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Combining Reactions to Calculate H H(kJ) N 2 + O 2 → 2NO 181 2NO + O 2 → 2NO 2 -113 Adding: N 2 + 2O 2 → 2NO 2 68 When two reactions add up to make a third combined reaction, we can also add the H for each reaction to calculate the H for the combined reaction. This is known as Hess’s Law, and is a consequence of the more fundamental Law of Conservation of Energy. H (kJ) C(graphite) + O 2 → CO 2 (g) -393.5 C(diamond) + O 2 → CO 2 (g) -395.4 Calculate H for C(graphite) → C(diamond) Combine the above reactions by first reversing the diamond combustion. H(kJ) CO 2 (g) → C(diamond) + O 2 +395.4 C(graphite) + O 2 → CO 2 -393.5 Adding: C(graphite) → C(diamond) +1.9 The above calculation indicates that there is very little energy difference between graphite and diamond. Does this mean that we can take 12.0 g (1 mol) of graphite, add a measly 1.9 kJ, (roughly equivalent to increasing the temperature of a cup of water by 2 degrees C) and instantly get 12.0 g diamond? Obviously not. Why not? H is a state function, and the difference of 1.9 kJ between diamond and graphite accurately represents the overall enthalpy difference between final and initial states. But H does not predict how easy the process is to accomplish or how fast the reaction will occur. This particular process (graphite → diamond), involves breaking down the strong bonds of graphite (lots of energy needed) and then reforming them as diamond (lots of energy released—all but 1.9 kJ of the energy required to break the graphite bonds). This process is difficult and expensive. The small value of H tells us nothing about the ease of actually accomplishing the change. H compares initial and final states, not the process of the change. The study of kinetics (chapter 14 of our text) looks at the factors that affect reaction rate as well as the mechanism and process of how chemical reactions occur.

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Given the following data: H(kJ) 2B + 2 3 O 2 → B 2 O 3 -1273 B 2 H 6 + 3O 2 → B 2 O 3 + 3H 2 O(g) -2035 H 2 + ½ O 2 → H 2 O( ) -286 H 2 O( ) → H 2 O(g) 44 Calculate H for 2B + 3H 2 → B 2 H 6 H(kJ) B 2 O 3 + 3H 2 O(g) → B 2 H 6 + 3O 2 +2035 Flip 2 nd equation to get B 2 H 6 on right side 2B + 2 3 O 2 → B 2 O 3 -1273 Keep 1 st eq. as is because we already have 2B on left side 3H 2 + 2 3 O 2 → 3H 2 O( ) -286 x 3 Multiply 3 rd eq by 3 to get 3H 2 on left side 3H 2 O( ) →3H 2 O(g) 44 x 3 Multiply 4 th eq. by 3 to cancel 3H 2 O( ) and 3H 2 O(g) Adding: 2B + 3H 2 → B 2 H 6 +36
Enthalpy of Formation Standard enthalpy of formation, designated as o f H , is defined as the H of a reaction in which one mole of a compound is formed from elements in their standard states, where the standard state is the state in which that element exists at 25 o C and 1.00 atm. o

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## This note was uploaded on 02/20/2012 for the course 160 161 taught by Professor Kim during the Fall '08 term at Rutgers.

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Lecture 16sf - Combining Reactions to Calculate H N2 O2 2NO...

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