### Bomb Calorimetry Calculation ExampleSuppose 0.100 mol of a metal (M) is combusted in excess \( \text{O}_2(g) \) in a bomb calorimeter:\[ 4\text{M}(s) + 3\text{O}_2 \rightarrow 2\text{M}_2\text{O}_3(s) \]The heat capacity of the empty calorimeter is 458 J/°C, and the calorimeter contains 450 mL of water. As a result of the reaction, the temperature of the water rises from 25.0°C to 40.8°C. Calculate \( \Delta E \) for the reaction in KJ/mol\(_{\text{rxn}}\). Assume:\[ D_{\text{H}_2\text{O}} = 1.00 \, \frac{g}{mL} \]\[ S_{\text{H}_2\text{O}} = 4.18 \, \frac{J}{g \cdot °C} \]### Possible Answers- \(-1.48 \times 10^3 \, \text{KJ/mol}\)- \(+1.48 \times 10^3 \, \text{KJ/mol}\)- \(+3.35 \times 10^3 \, \text{KJ/mol}\)- \(-3.35 \times 10^3 \, \text{KJ/mol}\) ### ExplanationThe aim is to calculate the change in internal energy (\( \Delta E \)) for the reaction based on the temperature change, the heat capacity of the calorimeter, and the properties of water. This involves understanding the principles of calorimetry, which measures the heat of chemical reactions or physical changes.

Given: 4M + 3O2 -----> 2M2O Heat released by 0.1 mol metal on reaction = heat gained by bomb calorimeter system = (450g x 4.18J/g/°C + 458J/°C) x (40.8-25)°C = 36956.2 J = 36.956 kJ Heat realeased by 4 mol metal = 36.956 x4 ÷0.1 or, ∆Erxn = - 1478.25 kJ/mol of reaction (heat is released so it is negative) ∆Erxn = = - 1.48 x 103 kJ/mol

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Suppose 0.100 mol of a metal (M) is combusted in excess O2(g) in a bomb calorimeter. 4M(s) + 302 2M203(s) The heat capacity of the empty calorimeter is 458 J/°C and the calorimeter contains 450 mL of water. As a result of the reaction, the temperature of the water rises from 25.0°C to 40.8°C Calculate AE for the reaction in KJ/molpxn: Assume, DH,0 = 1.00 _ mL = 4.18 g.ºC O -1.48 X 103 KJ/mol O +1.48 X 10³ KJ/mol O +3.35 X 10³ KJ/mol O -3.35 X 103 KJ/mol

Suppose 0.100 mol of a metal (M) is combusted in excess O2(g) in a bomb calorimeter. 4M(s) + 302 2M203(s) The heat capacity of the empty calorimeter is 458 J/°C and the calorimeter contains 450 mL of water. As a result of the reaction, the temperature of the water rises from 25.0°C to 40.8°C Calculate AE for the reaction in KJ/molpxn: Assume, DH,0 = 1.00 _ mL = 4.18 g.ºC O -1.48 X 103 KJ/mol O +1.48 X 10³ KJ/mol O +3.35 X 10³ KJ/mol O -3.35 X 103 KJ/mol

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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