Delta Go = -RTlnKeq
delta Go = 5.95 kJ/mole = 5.95 X 1000 = 5950 J/mole ( 1 kj = 1000 J )
putting the values and finding Keq
5950 = -8.314 X 298 X ln Keq
ln Keq = -5950 / 2477.572 = -2.4015
Keq = e^-2.402 = 0.0905
suppose the equilibrium reaction is :-
chair 1 <--------> chair 2
now as Keq is less than 1 ....so chair 1 will be more stable
Keq = [chair2]/[chair 1 ] = 0.0905
this means that [chair 2] ~ 0.0905 and [chair 1] ~ 1
[total] = [chair 2] + [chair 1] ~ 1 + 0.0905= 1.0905
percentage of chair 1 = [chair 1] / [total] = 1 / 1.0905 X 100 = 91.70 %
Answer:
2.22 g/L
Explanation:
There's a relationship using the ideal gas law between molar mass and density:
, where MM is the molar mass, d is the density, R is the gas constant, T is the temperature, and P is the pressure.
We know from the problem that MM = 32.49 g/mol, T = 458 Kelvin, and P = 2.569 atm. The gas constant, R, in terms of the units atm and Kelvin is 0.08206. Let's substitute these values into the formula:


Solve for d:
d * 0.08206 * 458 K = 32.49 * 2.569
d = (32.49 * 2.569) / (0.08206 * 458 K) ≈ 2.22 g/L
The answer is thus 2.22 g/L.
<em>~ an aesthetics lover</em>
the transition of a substance directly from the solid to the gas phase, without passing through the intermediate liquid phase.