We are asked to find the value of ΔG°rxn from the equilibrium concentrations of the reactants and products. We can use the following formula:
ΔG°rxn = -RTlnK
The value of R = 8.314 J/Kmol, T = 298.15 K and we are given the equilibrium constant Keq = 2.82.
The question provides equilibrium concentrations and then asks to find ΔG°rxn when more of a product is added to the reaction mixture. However, you are asked to find ΔG after the reaction has settled down and reached equilibrium once more. Therefore, we can simply use Keq = 2.82 still and solve for ΔG.
ΔG°rxn = -(8.314 J/Kmol)(298.15 K)(ln(2.82))
ΔG°rxn = -2570 J/mol
ΔG°rxn = -2.57 kJ/mol
Under equilibrium conditions at standard temperature and pressures, the value of ΔG°rxn = -2.57 kJ/mol.
Answer:
Q1) solids are strong, gasses are weak
Explanation:
- Solids are Tightly packed molecule with no gap or space in between.
- Gasses are loosely packed and move freely in our atmosphere.
- Examples of solids are Wood, steel, plastic etc
- Example of gasses are oxygen, Carbon dioxide, nitrogen
Please mark me as brainliest
Thank you :-)
Maybe to not get rained on.
Hahhahahaha I ain't sure tho
ΔG⁰ = ΔH⁰ - TΔS
ΔH⁰ = Hf,(CH₃OH) - Hf,(CO) = -238.7 + 110.5 = -128.2 kJ/mol
ΔS = S(CH₃OH) - S(CO) - 2S(H₂) = 126.8 - 197.7 - 2 x 130.6 = -332.1 J/mol.K
So
ΔG⁰ = - 128200 + 332.1 T
For the reaction to be spontaneous:
ΔG⁰ < 0
So: -128200 + 332.1 T < 0
332.1 T < 128200
T < 386.028 K