Question

The equilibrium constant for a reaction is 0.48 at 25 °C. What is the value of ΔG° (kJ/mol) at this temperature?
(A) 1.8
(B) -4.2
(C) 1.5 x 10^2
(D) 4.2
(E) More information is needed.

Answer 1

Answer:

a) 1.8 kJ

$1.8x10^{3} J$

Explanation:

ΔG=-RtLnK

Just input the data into the equation to gives us the result, in Joules.

ΔG=-8.314*(25+273.15)*ln(0.48)

ΔG=1819,37 J

But there is only 2 significant figures, so we express it as $1.8x10^{3}J$ or 1.8 kJ

Answer 2

The value of ΔG° (kJ/mol) at a temperature of 25°C is 1.8kJmol.

How to calculate gibbs free energy (∆G°)?

Gibbs free energy can be calculated using the following formula:

ΔG = -RtlnK

Where;

• R = gas law constant (8.314)
• T = temperature in Kelvin
• K = equilibrium constant

ΔG = -8.314 (298) × ln(0.48)

∆G = 1819.37

ΔG= 1.8KJmol

Therefore, the value of ΔG° (kJ/mol) at a temperature of 25°C is 1.8kJmol.

Learn more about Gibb free energy at: https://brainly.com/question/9552459