Question

Okay I did lab of the decomposition of H202 using the catalyst Fe(NO3)3.... I have to calculate the % error using this Equation: I will post the rest below

Answer 1

\[%= (literature value for heat of decomposition, kJ/mol) - (experimentally determined heat of decomposition , kJ/mol) \div (literature value for heat of decomposition, kJ/ mol ) x 100\]

the literature value is given which is -94.6 kJ/mol of H202... my question is how do i find the experimentally determined heat of decomposition?

Answer 2

\[%= (literature value for heat of decomposition, kJ/mol) - (experimentally determined heat of decomposition , kJ/mol) / (literature value for heat of decomposition, kJ/ mol ) x 100\]


the literature value is given which is -94.6 kJ/mol of H202... my question is how do i find the experimentally determined heat of decomposition?

Answer 3

Did you do the experiment in a calorimeter?

Answer 4

yes..but we used a Styrofoam cup instead

Answer 5

also are you able to view my equation?...its not showing for me and i posted it twice

Answer 6

No. Not showing up here.

You should be able to measure the heat released (or absorbed) during the reaction by measuring the change in temperature of the water.

Answer 7

also the concentration of the H202 is 0.8818 M .okay i will try to post the equation again..one second

Answer 8

( literature value for heat of decomposition, kJ/mol) - ( experimentally determined heat of decomposition kJ/mol) / ( literature value for heat of decomposition, kJ/mol) x 100


and the literature is -94.6 kJ/mol of H202...i just don't know how to find the experientially determined heat

Answer 9

DId you measure the temperature of the solution in the calorimeter directly?

Answer 10

yes with a thermometer....actually is there way to find the density of 0.8818 M of H202 i think i can figure out the experimentally determined heat if i had that?

Answer 11

eashmore are you still there?

Answer 12

Sorry. I got engrossed with my homework.

If we know the change in temperature of the solution in the calorimeter, we can easily find the heat released.

If you are given the specific heats of the H2O2 and Iron nitrate, we can account for them, but this would require understanding of reacting flows, which I doubt you've covered.

Let's treat it as pure water (unless your lab manual states otherwise).

The heat of the reaction can be expressed as\[Q = m_{water} c_{water}\Delta T\]

Answer 13

hey actually i just found the missing information to help me solve this equation now...it was hidden in the lab manual...hahah....sorry to waster your time...thanks for you help though!

Answer 14

Not a problem. Got you got it figured out.