Please Evaluate my answer (I feel im doing something wrong)

  • Thread starter Frank665
  • Start date
In summary, the enthalpy change for the reaction between 5.61 g of KOH and an unknown reactant, with an initial temperature of 20.0 oC and a final temperature of 24.1 oC, is 961.4 J/mol. This was calculated using the experimental values for heat and the specific heat capacity of water.
  • #1
Frank665
2
0
Question: Use the experimental values to calculate the enthalpy change in each system.

Observation
Quantity of reactant 1 5.61 g KOH(s)
Quantity of reactant 2 N/A
Initial Temperature (oC) 20.0
Final Temperature (oC) 24.1


MY ANSWER



ΔT = tf - ti
= 24.1 oC - 20.0 oC
= 4.1 oC

qt = 0

qt = qreaction+ qwater

qreaction= - qwater

qp= - mwaterCp(water) DT

qp = mc ΔT

= (5.61 g) (4.186 joule/gram °C) (4.1 oC)

=96.14 J

Where H is the Hfinal minus Hinitial and q is heat
ΔH = ( mcΔT)/n

n = (mass of compound)/(molar mass of compound)


= (5.61 g KOH(s))/(56.1 g/mol)

=
 
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  • #2
0.1 mol

ΔH = (96.14 J)/(0.1 mol)

= 961.4 J/mol

Therefore, the enthalpy change for the reaction is 961.4 J/mol.
 
  • #3
0.1 mol KOH(s)

ΔH = (96.14 J)/(0.1 mol)

= 961.4 J/mol

Thank you for providing your answer. However, in order for me to evaluate your answer, I would need more information about the experimental setup and the reaction being studied. It is important to include the units for all values, as well as the specific heat capacity of water (4.186 J/g°C) and the molar mass of KOH (56.1 g/mol). Additionally, it is important to check your calculations for accuracy and make sure all units cancel out to give the correct units for enthalpy change (J/mol). Overall, your approach seems correct, but without more context and information, I cannot fully evaluate your answer.
 

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