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

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In summary, the addition of hydrochloric acid to a silver nitrate solution causes a reaction that produces silver chloride and changes the temperature of the solution. Using the given information, the enthalpy change for the reaction was calculated to be -0.5752 kJ, rounded to two significant figures.
  • #1
sp3sp2sp
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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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  • #2
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:
 
  • #3
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:
  • #4
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?
 

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

1. What is Delta-H for a reaction?

Delta-H for a reaction, also known as enthalpy change, is the difference in energy between the products and the reactants of a chemical reaction. It represents the overall heat absorbed or released during the reaction.

2. Why is Delta-H important?

Delta-H is important because it helps us understand the thermodynamics of a chemical reaction. It tells us whether the reaction is exothermic (releases heat) or endothermic (absorbs heat), and how much energy is involved in the reaction.

3. How is Delta-H measured?

Delta-H can be measured experimentally using a calorimeter, which measures the change in temperature of the reactants and products. It can also be calculated using Hess's Law, which states that the enthalpy change of a reaction is equal to the sum of the enthalpy changes of the individual steps of the reaction.

4. What factors affect the value of Delta-H?

The value of Delta-H is affected by the type of chemical reaction, the concentrations of the reactants and products, and the temperature and pressure at which the reaction occurs. Catalysts can also affect the value of Delta-H by lowering the activation energy of the reaction.

5. How does Delta-H relate to the spontaneity of a reaction?

Delta-H is related to the spontaneity of a reaction through the Gibbs free energy equation: ΔG = ΔH - TΔS. A negative value for Delta-H means that the reaction releases heat and is more likely to occur spontaneously. However, the temperature and entropy changes also play a role in determining the spontaneity of a reaction.

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