Answer : The concentration of HI (g) at equilibrium is, 0.643 M
Explanation :
The given chemical reaction is:
Initial conc. 0.10 0.10 0.50
At eqm. (0.10-x) (0.10-x) (0.50+2x)
As we are given:
The expression for equilibrium constant is:
![K_c=\frac{[HI]^2}{[H_2][I_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BHI%5D%5E2%7D%7B%5BH_2%5D%5BI_2%5D%7D)
Now put all the given values in this expression, we get:
x = 0.0713 and x = 0.134
We are neglecting value of x = 0.134 because the equilibrium concentration can not be more than initial concentration.
Thus, we are taking value of x = 0.0713
The concentration of HI (g) at equilibrium = (0.50+2x) = [0.50+2(0.0713)] = 0.643 M
Thus, the concentration of HI (g) at equilibrium is, 0.643 M
Worms are fully developed insects and larvae are not.
A first order reaction, with a half-life of 125 s, has 1/16 of the original amount left after 500 seconds.
<h3>What is a first order reaction?</h3>
It is a chemical reaction in which the rate of reaction is directly proportional to the concentration of the reacting substance.
First, we will calculate the rate constant using the following expression.
ln ([A]/[A]₀) = - k × t
ln (1/16[A]₀/[A]₀) = - k × 500 s
k = 5.55 × 10⁻³ s⁻¹
where,
- [A] is the final concentration of the reactant.
- [A]₀ is the initial concentration of the reactant.
- k is the rate constant.
- t is the elapsed time.
Next, we can calculate the half-life (th) using the following expression.
th = ln 2 / k = ln 2 / (5.55 × 10⁻³ s⁻¹) = 125 s
A first order reaction, with a half-life of 125 s, has 1/16 of the original amount left after 500 seconds.
Learn more about first order reactions here: brainly.com/question/518682
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Answer:
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