Answer:
The equilibrium concentration of NO is 0.001335 M
Explanation:
Step 1: Data given
The equilibrium constant Kc is 0.0025 at 2127 °C
An equilibrium mixture contains 0.023M N2 and 0.031 M O2,
Step 2: The balanced equation
N2(g) + O2(g) ↔ 2NO(g)
Step 3: Concentration at the equilibrium
[N2] = 0.023 M
[O2] = 0.031 M
Kc = 0.0025 = [NO]² / [N2][O2]
Kc = 0.0025 = [NO]² / (0.023)(0.031)
[NO] = 0.001335 M
The equilibrium concentration of NO is 0.001335 M
Answer:
I think the answer is true, sorry if I am wrong
Explanation:
Answer:
If mass increases, force increases.
Explanation:
hope this helps, pls mark brainliest :D
Answer:
2 moles for the first 2 questions and 100g of water
Explanation:
Answer:
K = 8.1 x 10⁻³
Explanation:
We are told here that these gas phase reactions are both elementary processes, thus the reactions forward and reverse are both first order:
A→B Rate(forward) = k(forward) x [A]
and for
B→A Rate(reverse) = k(reverse) x [B]
At equilibrium we know the rates of the forward and reverse reaction are equal, so
k(forward) x [A] = k(reverse) x [B] for A(g)⇌B(g)
⇒ k(forward) / k(reverse) = [B] / [A] = K
4.7 x 10⁻³ s⁻1 / 5.8 x 10⁻¹ s⁻¹ = 8.1 x 10⁻³ = K
Notice how this answer is logical : the rate of the reverse reaction is greater than the forward reaction ( a factor of approximately 120 times) , and will be expecting a number for the equilibrium constant, K, smaller than one where the reactant concentration, [A], will prevail.
It is worth to mention that this is only valid for reactions which are single, elementary processes and not true for other equilibria.