Calculating Delta-H for a Reaction using a Coffee-Cup Calorimeter Method

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Discussion Overview

The discussion revolves around calculating the enthalpy change (ΔH) for the reaction between hydrochloric acid and silver nitrate using a coffee-cup calorimeter method. Participants are examining the calculations involved in determining ΔH based on temperature changes and solution concentrations, with a focus on the methodology and potential errors in the calculations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents a calculation for ΔHrxn based on the reaction and temperature change, initially arriving at a value of 580 J, which they later suspect is incorrect.
  • Another participant requests clarification on the calculation steps, specifically questioning the numbers used in the equation for heat (q).
  • A participant corrects their earlier calculation of the temperature change (ΔT) and recalculates the heat (q) using the correct ΔT of 0.81 °C, leading to a new value of -28.76 J for ΔHrxn per 0.05 mol of AgNO3.
  • There is a mention of needing to express the final answer to two significant figures, resulting in -0.58 kJ.
  • Another participant confirms the correct determination of moles of AgNO3 and prompts further exploration of the heat calculation for the total volume of liquid heated.

Areas of Agreement / Disagreement

Participants generally agree on the method of calculating ΔHrxn and the number of moles of AgNO3 involved. However, there are uncertainties regarding specific calculations and potential errors, indicating that the discussion remains unresolved.

Contextual Notes

Participants express uncertainty about specific values used in calculations, such as the temperature change and the resulting heat calculations. There is also a lack of consensus on the correctness of the final ΔHrxn value due to identified mistakes in earlier calculations.

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Homework Statement


The addition of hydrochloric acid to a silver nitrate solution precipitates silver chloride according to the reaction:

AgNO3(aq)+HCl(aq)→AgCl(s)+HNO3(aq)

When 500 mL of 0.100 MAgNO3 is combined with 500 mL of HCl in a coffee-cup calorimeter, the temperature changes from 23.40 ∘C to 24.21 ∘C. Calculate ΔHrxnfor the reaction as written. Use 1.00 g/mL as the density of the solution and C=4.18J/(g⋅∘C) as the specific heat capacity

Homework Equations


MM AgNO3 = 169.88g.mol
q = mcdeltaT = deltaH_rxn

The Attempt at a Solution


q = [(500mL/1000)(0.100M)(169.88g/mol) ] * 4.18J/g/degC * 13.31degC = 28.75J = q =ΔHrxn for 0.05mol
28.75J * 20 = 580J = 0.58kJ (2 sig-fig)
= wrong
Thans for any help
 
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Please elaborate on

sp3sp2sp said:
q = [(500mL/1000)(0.100M)(169.88g/mol) ] * 4.18J/g/degC * 13.31degC

None of the numbers here makes sense to me :frown:
 
thanks for the reply. It is from q = m*c*deltaT. I did make mistake for delta T, which is corrected below.
First I calculated the grams of AgNO3:
[(500mL/1000)(0.100M) * (169.88g/mol) = 8.494g

I was provided C=4.18J/(g⋅∘C) as the specific heat capacity in question stem.

temp change is T-final - T-initial = 24.21 - 23.40 = 0.81degC

then q = [(500mL/1000)(0.100M)(169.88g/mol) ] * 4.18J/g/degC * 0.81degC = 28.76J

q = -enthalpy = -28.76J for .05mol of AgNO3.

20* .05mol = 1 mol of AgNO3

so 20 * -28.76J = -575.2J = -0.5752kJ

Answer needed to be to two sig figs, so = 0.58kJ

I know there's mistakes in this but I am not sure where they are...thanks again for any help
 
Last edited:
You determined that 0.05 moles of AgNO3 reacted. This is correct.

You had 1000 ml of liquid that were heated 0.81 C. How many joules of heat does this correspond to? How many joules per mole of AgNO3 does this correspond to?
 

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