Chemistry, 27.07.2019 21:00 Jorianes

Ineed to know identify a potential source of error that may have occurred during the calorimetry experiment. justify your response in two or more complete sentences.

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A source of error is any factor that may affect the outcome of an experiment. There are countless conceivable sources of error in any experiment; you want to focus on the factors that matter most. Identify each source of error specifically and then explain how that source of error would have affected the results. Keep in mind that an "error" to a scientist does not mean "mistake"; it more closely means "uncertainty".
Many students are tempted to say "human error", but this term is vague and lazy; any decent teacher will not accept it. Instead, think about specific things that happened during the lab exercise where the end results may have been affected.
To give an example one might find in a bio lab: perhaps a water bath's temperature was not monitored very carefully and you found that an enzyme's activity was greater than you expected. In that case, you could write something like,
"The temperature of the water bath during this exercise was not monitored carefully. It is possible that it was warmer or cooler than intended, and this would have affected the enzyme activity accordingly. The fact that our enzyme activity was found to be higher than expected leads me to believe that perhaps the water bath was too warm."

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

Answer from: Quest

the temperature of the gas, in k, not celsius.

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Ineed to know identify a potential source of error that may have occurred during the calorimetry ex...