Rate = k * [A]^2 * [B]^1
<span>Use the data from any trial to calculate k. </span>
<span>k = (rate)/([A]^2 * [B]^1) </span>
<span>E.g., for Trial 1, we have </span>
<span>rate = 3.0×10−3 M/s </span>
<span>[A] = 0.50 M </span>
<span>[B] = 0.010 M </span>
<span>Plug those numbers in and crank out the answer. </span>
<span>Now with the calculated value of k, calculate the initial rate for [A] = 0.50 M and [B] = 0.075 M </span>
<span>rate = k * [A]^2 * [B]^1 </span>
<span>k = calculated value </span>
<span>[A] = 0.50 M </span>
<span>[B] = 0.075 M</span>
Ok boomer..................
51: The only coefficient is in the first one i think and it would be a 3
52: Neon (Ne)
53: Mg has a low nuclear charge so all it's electrons would be lost in the electron shell, making the ion smaller than the atom.
Answer : The value of equilibrium constant for this reaction at 328.0 K is
Explanation :
As we know that,
where,
= standard Gibbs free energy = ?
= standard enthalpy = 151.2 kJ = 151200 J
= standard entropy = 169.4 J/K
T = temperature of reaction = 328.0 K
Now put all the given values in the above formula, we get:
The relation between the equilibrium constant and standard Gibbs free energy is:
where,
= standard Gibbs free energy = 95636.8 J
R = gas constant = 8.314 J/K.mol
T = temperature = 328.0 K
K = equilibrium constant = ?
Now put all the given values in the above formula, we get:
Therefore, the value of equilibrium constant for this reaction at 328.0 K is
The answer is potassium magnate