When in danger, bombardier beetles can fire a hot, toxic mixture of chemicals at their attacker. This mixture contains quinone, C6H4O2, a compound that is formed by the reaction of hydroquinone, CH4(OH)2, with hydrogen peroxide, H2O2. The equation for the overall reaction is: CH4(OH)2(aq) + H2O2(aq) → C6H4O2(aq) + 2H2O(ℓ) Use the following data to calculate the enthalpy change, in kJ mol-1 , for the above reaction. CH4(OH)2(aq) → C6H4O2 (aq) + H2(g) ΔH = +177.4 kJ mol−1 H2(g) + O2(g) → H2O2(aq) ΔH = −191.2 kJ mol−1 H2(g) + ½O2(g) → H2O(g) ΔH = −241.8 kJ mol−1 H2O(g) → H2O(ℓ) ΔH = −43.8 kJ mol−1
When in danger, bombardier beetles can fire a hot, toxic mixture of chemicals at their attacker. This mixture contains quinone, C6H4O2, a compound that is formed by the reaction of hydroquinone, CH4(OH)2, with hydrogen peroxide, H2O2. The equation for the overall reaction is: CH4(OH)2(aq) + H2O2(aq) → C6H4O2(aq) + 2H2O(ℓ) Use the following data to calculate the enthalpy change, in kJ mol-1 , for the above reaction. CH4(OH)2(aq) → C6H4O2 (aq) + H2(g) ΔH = +177.4 kJ mol−1 H2(g) + O2(g) → H2O2(aq) ΔH = −191.2 kJ mol−1 H2(g) + ½O2(g) → H2O(g) ΔH = −241.8 kJ mol−1 H2O(g) → H2O(ℓ) ΔH = −43.8 kJ mol−1
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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When in danger, bombardier beetles can fire a hot, toxic mixture of chemicals at their attacker. This
mixture contains quinone, C6H4O2, a compound that is formed by the reaction of hydroquinone, CH4(OH)2,
with hydrogen peroxide, H2O2. The equation for the overall reaction is:
CH4(OH)2(aq) + H2O2(aq) → C6H4O2(aq) + 2H2O(ℓ)
Use the following data to calculate the enthalpy change, in kJ mol-1
, for the above reaction.
CH4(OH)2(aq) → C6H4O2 (aq) + H2(g) ΔH = +177.4 kJ mol−1
H2(g) + O2(g) → H2O2(aq) ΔH = −191.2 kJ mol−1
H2(g) + ½O2(g) → H2O(g) ΔH = −241.8 kJ mol−1
H2O(g) → H2O(ℓ) ΔH = −43.8 kJ mol−1
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