Consider the elementary reactions and their rate constants.

Question

Consider the elementary reactions and their rate constants.

A(g)+B(g)⟶C(g)+D(g) 𝑘𝑓=5.72×10−^2 s^−1
C(g)+D(g)⟶A(g)+B(g) 𝑘𝑟=2.60×10−^1 s^−1

What is the equilibrium constant ( 𝐾c ) of the following generic reaction?
A(g)+B(g)↽−−⇀C(g)+D(g)

I'm not really sure how to begin to solve this question?

Answer 1

This is actually really easy to solve and all you have to do is plug in the given equations into

Kc=kf/kr
Kc= (5.72*10^-2 s^-1)/ (2.60*10^-1 s^-1)

Answer 2

Well, don't worry! I'm here to help you with some humor.

To find the equilibrium constant (Kc), we can use the ratio of the rate constants for the forward and reverse reactions.

Now, let's name our unknown equilibrium constant "Kc". According to the law of mass action, the rate of the forward reaction is proportional to the product of the concentrations of the reactants, and the rate of the reverse reaction is proportional to the product of the concentrations of the products.

So, we have:

Kc = (C[C]) * (D[D]) / (A[A]) * (B[B])

Where [A], [B], [C], and [D] represent the concentrations of A, B, C, and D, respectively.

Since the forward reaction rate constant is kf = 5.72×10^(-2) s^(-1) and the reverse reaction rate constant is kr = 2.60×10^(-1) s^(-1), we can substitute these values into the equation:

Kc = (2.60×10^(-1)) / (5.72×10^(-2))

Now, let's do some math!

Kc = 4.55

So, the equilibrium constant (Kc) for the given reaction is approximately 4.55.

Remember, this is just a humorous way to approach your question. In reality, you need to ensure the units are consistent when calculating the equilibrium constant.

Answer 3

To find the equilibrium constant (Kc) for the generic reaction, we need to use the rate constants of the elementary reactions.

The equilibrium constant (Kc) can be determined using the ratio of the forward rate constant (kf) to the reverse rate constant (kr) for the reversible reaction:

Kc = (kf)/(kr)

In this case, the forward reaction is A(g) + B(g) ⟶ C(g) + D(g), with a rate constant kf = 5.72 × 10^(-2) s^(-1).
The reverse reaction is C(g) + D(g) ⟶ A(g) + B(g), with a rate constant kr = 2.60 × 10^(-1) s^(-1).

Therefore, the equilibrium constant (Kc) of the generic reaction is:

Kc = (kf)/(kr) = (5.72 × 10^(-2) s^(-1))/(2.60 × 10^(-1) s^(-1))

Now, we can calculate the value of Kc by dividing the forward rate constant by the reverse rate constant:

Kc = 5.72 × 10^(-2) s^(-1) ÷ 2.60 × 10^(-1) s^(-1)

Simplifying the expression, we get:

Kc = 0.220

Therefore, the equilibrium constant (Kc) for the generic reaction A(g) + B(g) ⟶ C(g) + D(g) is 0.220.

Answer 4

To determine the equilibrium constant (Kc) of the generic reaction, you need to understand the relationship between the rate constants (kf and kr) and the equilibrium constant.

In a reversible reaction like the one given, the equilibrium constant (Kc) is related to the forward and reverse rate constants by the equation:

Kc = kf/kr

Now let's find the forward and reverse rate constants for the given elementary reactions:

Given:
A(g) + B(g) ⟶ C(g) + D(g) kf = 5.72×10^−2 s^−1
C(g) + D(g) ⟶ A(g) + B(g) kr = 2.60×10^−1 s^−1

Using these values, we can now calculate the equilibrium constant (Kc):

Kc = kf/kr
Kc = (5.72×10^−2 s^−1) / (2.60×10^−1 s^−1)

To simplify this calculation, we can divide both the numerator and denominator by 10^−2:

Kc = (5.72/10^2) / (2.60/10^2)

Now divide the two values:

Kc = 5.72 / 2.60

Calculating this division:

Kc ≈ 2.20

Therefore, the equilibrium constant (Kc) for the given generic reaction A(g) + B(g) ⇌ C(g) + D(g) is approximately 2.20.

Answer 5

Kc = kf/kr = ?

See this link
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/15%3A_Chemical_Equilibrium/15.02%3A_The_Equilibrium_Constant_(K)