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.
The scientific evidence that scientists use in supporting or critiquing the conclusions of experiments usually consists observations based on large sample sizes.
Volume:
2.00 x 11.0 x 11.0 => 242 cm³
mass : 213 g
D = m / V
D = 213 / 242
D = 0.880 g/cm³
Answer B
hope this helps!
Answer:
C) Q < K, reaction will make more products
Explanation:
- 1/8 S8(s) + 3 F2(g) ↔ SF6(g)
∴ Kc = 0.425 = [ SF6 ] / [ F2 ]³
∴ Q = [ SF6 ] / [ F2 ]³
∴ [ SF6 ] = 2 mol/L
∴ [ F2 ] = 2 mol/L
⇒ Q = ( 2 ) / ( 2³)
⇒ Q = 0.25
⇒ Q < K, reaction will make more products
