Question

The value of ΔGo for the precipitation reaction,

Answer 1

Answer:

$\Delta G=-44.9kJ$

Explanation:

We know, $\Delta G=\Delta G^{0}+RTlnQ$

where Q is the reaction quotient and for this reaction, it is expressed as-

                        $Q=\frac{1}{[Ca^{2+}][CO_{3}^{2-}]}$            

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, $\Delta G=(-48.1\times 10^{3}J)+(8.314\frac{J}{K}\times 298K)ln(\frac{1}{0.312\times 0.898})$

or, $\Delta G=-44948J$

or, $\Delta G=-44.9kJ$