Answer:
It takes 1.32x10⁻⁷s for the concentration of A to fall by a factor of 8
Explanation:
The equation that represents a first-order kinetics is:
Ln ([A] / [A]₀] = -kt
<em>Where [A] is actual concentration, [A]₀ is initial concentration, K is rate constant (For the given problem, 1.57x10⁷s⁻¹ and t is time.</em>
<em />
As you want the time when you have [A] in a factor of 8 = [A] / [A]₀ = 1/8
Replacing:
Ln ([A] / [A]₀] = -kt
Ln (1/8) = -1.57x10⁷s⁻¹*t
t = 1.32x10⁻⁷s
<h3>It takes 1.32x10⁻⁷s for the concentration of A to fall by a factor of 8</h3>
Answer:
The pressures will remain at the same value.
Explanation:
A catalyst is a substance that alter the rate of a chemical reaction. It either speeds up the or slows down the rate of a chemical reaction.
While a catalyst affects the rate, it is noteworthy that it has no effect on the equilibrium position of the chemical reaction. A catalyst works by creating an alternative pathway for the reaction to proceed. Most times, it decreases the activation energy needed to kickstart the chemical reaction.
Hence, we know that it has no effect on the equilibrium position. Factors affecting equilibrium position includes, temperature and concentration of reactants and products( pressure in terms of gases).
The reactants and the products here are gaseous, and as such pressure affects the equilibrium position. Now, we have established that the equilibrium position is unaffected. And as such the pressure affecting it does not change.
Thus, we have established that the pressure of the products and reactants are unaffected and as such they remain at their value unaffected.
There has to be something else then just what you said
Me neither tbh Goodluck with that tho
