In a coffee-cup calorimeter, 50.0ml of 0.100M AgNO3 and50.0ml of 0.100M HCl are mixed to yield the following reaction:
Ag+(aq) + Cl-(ag) >AgCl(s)
The two solutions were initially at 22.60oC, and the final temperature is 23.40oC. Calculate the heat that accompanies this reaction in KJ/mol of AgCl formed. Assume that the combined solution has a mass of 100.0 g and a specific heat capacity of 4.18 J/oC.g.

Where q is heat of reaction, C is specific heat capacity (4.18J/°Cg), m is mass of solution (100.0g) and ΔT is temperature change (23.40°C-22.60°C = 0.80°C)

Replacing:

q = -4.18J/°Cg×100.0g×0.80°C

q = -334.4J

Now, in the reaction:

Ag⁺ + Cl⁻→ AgCl

AgNO₃ as source of Ag⁺ and HCl as source of Cl⁻

Moles that react are:

0.050L× (0.100mol /L) = 0.0050moles

If 0.0050 moles produce -334.4J. Heat of reaction is:

-334.4J / 0.0050moles = -66880J/mol = -66.88KJ/mol

andromache andromache · Answer 2 · 2022-03-24T17:01:26+01:00

The heat that accompanies this reaction in KJ/mol of AgCl formed should be -66.88KJ/mol.

Calculation of the heat:

here we used a calorimeter formula is

q = -C×m×ΔT

Here

q is the heat of reaction,

C is specific heat capacity (4.18J/°Cg),

m is mass of solution (100.0g)

and ΔT is temperature change (23.40°C-22.60°C = 0.80°C)

So,

q = -4.18J/°Cg×100.0g×0.80°C

q = -334.4J

Now, in the reaction:

Ag⁺ + Cl⁻→ AgCl

AgNO₃ as a source of Ag⁺ and HCl as a source of Cl⁻

Now

Moles that react are:

0.050L× (0.100mol /L) = 0.0050moles

In the case when 0.0050 moles produce -334.4J.

So, the Heat of reaction is:

= -334.4J / 0.0050moles

= -66880J/mol

= -66.88KJ/mol

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