Respuesta :
Answer:
[tex]\Delta G=-44.9kJ[/tex]
Explanation:
We know, [tex]\Delta G=\Delta G^{0}+RTlnQ[/tex]
where Q is the reaction quotient and for this reaction, it is expressed as-
[tex]Q=\frac{1}{[Ca^{2+}][CO_{3}^{2-}]}[/tex]
where species under third bracket represent concentrations in molarity.
T represents temperature in kelvin scale
Here T = 298 K
R = gas constant = 8.314 J/K
So, [tex]\Delta G=(-48.1\times 10^{3}J)+(8.314\frac{J}{K}\times 298K)ln(\frac{1}{0.312\times 0.898})[/tex]
or, [tex]\Delta G=-44948J[/tex]
or, [tex]\Delta G=-44.9kJ[/tex]
The value of ΔG for the precipitation reaction is -44.948 kJ.
How we calculate the change in Gibb's free energy?
Chnage in Gibb's free energy of the reaction will be calculated as:
ΔG = ΔG° + RTlnQ, where
ΔG° = standard gibb's free energy = -48.1 kJ = -48100 kJ
R = universal gas constant = 8.314 J/k
T = temperature = 298 K
Q = ratio of the concentration of products to the reactants
Given chemical reaction is:
Ca²⁺(aq) + CO₃²⁻(aq) → CaCO₃(s)
Value of Q for the given reaction = 1 / [Ca²⁺][CO₃²⁻]
Given values of concentration of Ca²⁺ & CO₃²⁻ are 0.312 and 0.898 M respectively.
Now on putting all these values on the above equation, we get
ΔG = -48100 + (8.314 × 298) ln(1/ 0.312 × 0.898)
ΔG = -44948 J
ΔG = -44.948 kJ
Hence, value of ΔG is -44.948 kJ.
To know more about gibb's free energy, visit the below link:
https://brainly.com/question/14415025
