qV rU Constant Volume Calorimetry Measuring U Heats of combustion r U o

# Qv ru constant volume calorimetry measuring u heats

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qV = rU Constant Volume Calorimetry, Measuring U Heats of combustion (r U o combustion ) are measured using a device called a Bomb Calorimeter. A combustible sample is reacted with excess O2 The heat capacity of the bomb is constant. The heat of reaction is found by: Constant Volume Calorimetry, Measuring U rxn cal cal water bomb q C T C C C = - ×∆ = + Octane, the primary component of gasoline combusts by the reaction: C8H18( l ) + 25/2 O2(g)  8 CO2(g) + 9H2O( l ) A 1.00 g sample of octane is burned in a bomb calorimeter that contains 1.20 kg of water surrounding the bomb. The temperature of the water rises to 33.20 °C from 25.00 °C when the octane is reacted. If the heat capacity of the bomb is 837 J/°C, calculate the heat of reaction per mole of octane. q rxn = q water + q bomb Since the temperature of the water rose, the reaction must have been exothermic: Therefore one can write: Calculating Heat in an Exothermic Reaction q rxn = mwaterCwaterTwater + q bombTwater q water q bomb q RXN = – 4803 J or – 48.0 kJ Heat transferred per mole qV: Calculating Heat in an Exothermic Reaction - = RXN q 4.184J g C ° × × 3 1.20 10 g ( 29 33.20 25.00 C × - ° J 837 C + ° ( 29 33.20 25.00 C × - ° 3 48.0 kJ kJ 5.48 10 mol mol 1.00g 114.2g - = - × × The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction. Reactants Products rH = ? unknown! Intermediate Reaction rH1 known rH2 k nown rH1 + rH2 = rH The sum of the H’s in one direction must equal the sum in the other direction. What if the enthalpy changes through another path are know? Why? Because enthalpy is a state function… Path independent ! so we can write… Hess’s Law Forming CO2 can occur in a single step or in a two steps. ∆rHtotal is the same no matter which path is followed. Hess’s law & Energy Level Diagrams Notice that the path from reactants to products in the desired reaction goes through an “intermediate compound” in the given reactions. This means that the path for hydrogen and nitrogen to produce ammonia goes through hydrazine (N2H4). Therefore, the path to the enthalpy of the reaction must be a sum of the two given reactions! Hess’s Law Problem: Example: Determine the rH for the reaction: 3H2(g) + N2(g)  2NH3(g) rHo = ??? Given: (1) 2 H2( g ) + N2( g )  N2H4( g ) ∆rH°1 = +95.4 kJ (2) N2H4( g ) + H2( g )  2NH3( g ) ∆rH°2 = –187.6 kJ adding equations (1) & (2) yields: 2 H2( g ) + N2( g ) + N2H4( g ) +H2( g )  N2H4( g ) + 2NH3( g ) Hess’s Law Problem: Example: Determine the rH for the reaction: 3H2(g) + N2(g)  2NH3(g) rHo = ??? Given: (1) 2 H2( g ) + N2( g )  N2H4( g ) ∆rH°1 = +95.4 kJ (2) N2H4( g ) + H2( g )  2NH3( g ) ∆rH°2 = –187.6 kJ Hess’s Law Problem: Example: Determine the rH for the reaction: 3H2(g) + N2(g)  2NH3(g) rHo = ??? Given: (1) 2 H2( g ) + N2( g )  N2H4( g ) ∆rH°1 = +95.4 kJ (2) N2H4( g ) + H2( g )  2NH3( g ) ∆rH°2 = –187.6 kJ adding equations (1) & (2) yields: 2 H2( g ) + N2( g )+ N2H4( g ) + H2( g ) N2H4( g )+ 2NH3( g ) Look what happens… / / 3H2(g) + N2(g)  2NH3(g) Hess’s Law Problem: Example: Determine the rH for the reaction: 3H2(g) + N2(g)  2NH3(g) rHo = ???  #### You've reached the end of your free preview.

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• Fall '07
• Moran
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