If 50.0 mL of 0.88 M H2O2 and 10.0 mL of 0.50 M Fe(NO3)3 were combined and a temperature change of 8.03°C was observed, The specific heat of water is 4.18 J/(g * ∘C) Calculate the enthalpy of decomposition of hydrogen peroxide (ΔH) determined from the experimental data. Record your answer with the proper significant figures and include the correct sign if needed. Assume the density and specific heat of the solution are the same as that of water.
Fe(NO3)3 is catalyst.
Moles of Hydrogen peroxide (n)
= Molarity × volume
= 0.88 (mol/L) × ( 50.0×10-3) L
= 0.044 .
Total volume of solution = ( 50.0+10.0) mL = 60.0 mL
Assuming density is equal to water (1g/mL).
Then, mass = (60.0×1) = 60 g
qsolution = mass × specific heat × temperature change
= 60 (g) × 4.18 (J/goc) × 8.03 (0c)
= 2014 J.
Then, qrxn = - qsoln = - 2014 J
Or, enthalpy of decomposition
Hrxn
= qrxn/n
= -(2014/0.044) = - 45771 J
= -(45771/1000) KJ/mol
= - 45.8 KJ/mol. (3 significant figures) .
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