if the temperature of 20.0 L of gas were increased to twice the initial value, while the pressure is held constant what would be
1 answer:
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Answer:
The correct option is: (D) -2.4 kJ/mol
Explanation:
<u>Chemical reaction involved</u>: 2PG ↔ PEP
Given: The standard Gibb's free energy change: ΔG° = +1.7 kJ/mol
Temperature: T = 37° C = 37 + 273.15 = 310.15 K (∵ 0°C = 273.15K)
Gas constant: R = 8.314 J/(K·mol) = 8.314 × 10⁻³ kJ/(K·mol) (∵ 1 kJ = 1000 J)
Reactant concentration: 2PG = 0.5 mM
Product concentration: PEP = 0.1 mM
Reaction quotient:
<u>To find out the Gibb's free energy change at 37° C (310.15 K), we use the equation:</u>
<u>Therefore, the Gibb's free energy change at 37° C (310.15 K): </u><u>ΔG = (-2.45 kJ/mol)</u>