Question

Consider the reaction: CO2(g) + CCl4(g) ⇌ 2 COCl2(g) ΔG° = 46.9 kJ Under the following conditions at 25 oC: LaTeX: P_{CO_2}P C O 2= 0.459 atm, LaTeX: P_{CCl_4}P C C l 4= 0.984 atm, and LaTeX: P_{COCl_2}P C O C l 2= 0.653 atm, ΔG for the reaction is , and the forward the reaction is

Answer 1

Answer : The value of $\Delta G_{rxn}$ is, $-47.0kJ/mole$

Explanation :

The formula used for $\Delta G_{rxn}$ is:

$\Delta G_{rxn}=\Delta G^o+RT\ln K_p$   ............(1)

where,

$\Delta G_{rxn}$ = Gibbs free energy for the reaction

$\Delta G_^o$ =  standard Gibbs free energy  = 46.9 kJ

R = gas constant = 8.314 J/mole.K

T = temperature = $25^oC=273+25=298K$

$K_p$ = equilibrium constant

First we have to calculate the value of $K_p$.

The given balanced chemical reaction is,

$CO_2(g)+CCl_4(g)\rightarrow 2COCl_2(g)$

The expression for reaction quotient will be :

$K_p=\frac{(p_{COCl_2})^2}{(p_{CO_2})\times (p_{CCl_4})}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

$K_p=\frac{(0.653)^2}{(0.459)\times (0.984)}$

$K_p=0.944$

Now we have to calculate the value of $\Delta G_{rxn}$ by using relation (1).

$\Delta G_{rxn}=\Delta G^o+RT\ln K_p$

Now put all the given values in this formula, we get:

$\Delta G_{rxn}=-46.9kJ/mol+(8.314\times 10^{-3}kJ/mole.K)\times (298K)\ln (0.944)$

$\Delta G_{rxn}=-47.0kJ/mol$

Therefore, the value of $\Delta G_{rxn}$ is, $-47.0kJ/mole$