Answer:
-66.88KJ/mol
Explanation:
It is possible to obtain the heat involved in a reaction using a calorimeter. Formula is:
q = -C×m×ΔT
<em>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)</em>
Replacing:
q = -4.18J/°Cg×100.0g×0.80°C
q = -334.4J
Now, in the reaction:
Ag⁺ + Cl⁻→ AgCl
<em>AgNO₃ as source of Ag⁺ and HCl as source of Cl⁻</em>
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 = <em>-66.88KJ/mol</em>
Adaptation actually and also following control measures on how to avoid it from happening
ΔG° at 450. K is -198.86kJ/mol
The following is the relationship between ΔG°, ΔH, and ΔS°:
ΔH-T ΔS = ΔG
where ΔG represents the common Gibbs free energy.
the enthalpy change, ΔH
The temperature in kelvin is T.
Entropy change is ΔS.
ΔG° = -206 kJ/mol
ΔH° equals -220 kJ/mol
T = 298 K
Using the formula, we obtain:
-220kJ/mol -T ΔS° = -206kJ/mol
220 kJ/mol +206 kJ/mol =T ΔS°.
-T ΔS = 14 kJ/mol
for ΔS-14/298
ΔS=0.047 kJ/mol.K
450K for the temperature Completing a formula with values
ΔG° = (450K)(-0.047kJ/mol)-220kJ/mol
ΔG° = -220 kJ/mol + 21.14 kJ/mol.
ΔG°=198.86 kJ/mol
Learn more about ΔG° here:
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