Hydrogen peroxide is often used as an oxidizing agent in rocketfuel mixtures (mu
ID: 688347 • Letter: H
Question
Hydrogen peroxide is often used as an oxidizing agent in rocketfuel mixtures (much like thedemo!). The reaction below details the decomposition of hydrogenperoxide.
2 H2O2(l) 2 H2O(l) + O2(g) Hrxn = -196.1 kJ
a. How much heat is released when 732 g of peroxide isdecomposed?
b. If the peroxide started at 19.6°C, what is the finaltemperature of the system? (Assume a
“closed” system that does not interact with thesurroundings, and that peroxide has the
same specific heat capacity of water).
c. At that “final” temperature and 1 atm of pressure,what volume of oxygen gas is
generated? (hint: which moles do you need for the ideal gaslaw?)
Explanation / Answer
2 H2O2(l) 2 H2O(l) +O2(g) :Hrxn = -196.1kJMolar mass of H2O2 is = 2 * 1 + 2 * 16 = 34 g 2 * 34 g of H2O2 on decomposition the enthalphy change is196.1 KJ 732 g of H2O2 on decomposition the enthalphy changeis X KJ X = ( 196.1 * 732 ) / 2 * 34 = 2110.958 KJ We know that Q = mcdt Where Q = amout of heat = H rxn = 2110.958 KJ = 2110958 J m = mass of H2O2 = 732 g c = specific heat capacity = 4.186 J / g oC dt = temperature difference = t - 19.6 Plug the values we have dt = t-19.6 = 688.92 oC t = 708.52 o C Also PV = nRT P = Pressure = 1 atm T = Temperature = 708.52 + 273 = 981.52 K V = Volume = ? R = gas constant = 0.0821 L atm / mol - K n = No. of moles = 1 mole Plug the values we have V = 80.5828 L m = mass of H2O2 = 732 g c = specific heat capacity = 4.186 J / g oC dt = temperature difference = t - 19.6 Plug the values we have dt = t-19.6 = 688.92 oC t = 708.52 o C Also PV = nRT P = Pressure = 1 atm T = Temperature = 708.52 + 273 = 981.52 K V = Volume = ? R = gas constant = 0.0821 L atm / mol - K n = No. of moles = 1 mole Plug the values we have V = 80.5828 L
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