Question

solid ca(hco3)2 decomposes as
Ca(hco3)2(s) --> caco3(s)+co2(g)+H2O(g)
if the total pressure is 0.2 bar at 420k what is the standard gibbs free energy change for the reaction?
SOMEONE ANSWER PROPERLY & FAST PLEASE !!!!

Answer 1

Answer:

Explanation:

Since CO2 and H2O are the only gases present in the equation:

Ca(HCO3)2-->CaCO3+CO2+H2O

Initial 1 - - -

Equil 1-x x x x

2x=0.2

=>x=0.1

Total number of moles=1+2x

Therefore Kp=(x/(1+2x))(x/(1+2x))0.2

=.0013 atm^2

Now using Kp=Kc(RT)^delta n

Delta n=2

Therefore Kc=Kp/(RT)^2

=1.093x(10^-6)

Using G=-2.303RTlog(Kc)

=47940.16Joules

=47.94kJ

Answer 2

Answer:

The Gibbs free energy change for the reaction is 47940.16J.

Explanation:

For the given reaction: Ca(HCO₃)₂(s) --> CaCO₃(s)+CO₂(g)+H₂O(g)

At equilibrium, 2x=0.2

∴x=0.1

The number of moles=1+2x

We know that,

Kp can be calculated by,

Kp$=\frac{x}{1+2x}\frac{x}{1+2x}0.2$

    =0.0013

Now substituting Kp=0.013 and n=2 in the formula $Kc=Kp(RT)^{n}$

∴Kc=1.093*10⁻⁶

We know that,

$G=-2.303RTlogKc$

    =47940.16J

Therefore, the Gibbs free energy change for the reaction is 47940.16J.