If the equilibrium constant for the conversion of isobutane to n-butane is 2.5 what is the value of the equilibrium constant for the reverese reaction?
If Keq = 2.5, then the reverse equation, it is Keq' = 1/2.5.
Well, finding the value of the equilibrium constant for the reverse reaction is as tricky as finding a clown in a library. But fear not, I'm here to spread some laughter and knowledge!
In this case, if the equilibrium constant for the conversion of isobutane to n-butane is 2.5, then the equilibrium constant for the reverse reaction (the conversion of n-butane to isobutane) can be found by taking the reciprocal of the original constant. So, 1 divided by 2.5 gives us a value of 0.4.
Therefore, the value of the equilibrium constant for the reverse reaction is 0.4. Just remember, when it comes to humor and chemical reactions, sometimes it's all about flipping things around!
To find the value of the equilibrium constant for the reverse reaction, we can use the relationship between the equilibrium constants of a forward and reverse reaction.
The equilibrium constant (K) for a chemical reaction is the ratio of the concentrations of products to the concentrations of reactants, each raised to their respective stoichiometric coefficients.
For the forward reaction:
isobutane → n-butane
And for the reverse reaction:
n-butane → isobutane
The equilibrium constant for the reverse reaction can be determined by taking the inverse (reciprocal) of the equilibrium constant for the forward reaction.
Therefore, to calculate the equilibrium constant for the reverse reaction, we can use the equation:
K(reverse) = 1/K(forward)
Substituting the given value of K(forward) = 2.5 into the equation:
K(reverse) = 1/2.5
K(reverse) = 0.4
Hence, the value of the equilibrium constant for the reverse reaction is 0.4.
To find the value of the equilibrium constant for the reverse reaction, you can use the relationship between the equilibrium constant of a reaction and the equilibrium constant of its reverse reaction.
For a reaction: aA + bB ⇌ cC + dD
The equilibrium constant, K, is given by the equation:
K = ([C]^c * [D]^d) / ([A]^a * [B]^b)
Where [A], [B], [C], and [D] represent the molar concentrations of the species.
In the given reaction, the conversion of isobutane (C4H10) to n-butane (C4H10) can be represented as:
isobutane ⇌ n-butane
The equilibrium constant is given as 2.5.
Let's assign the stoichiometric coefficients for the reaction:
isobutane ⇌ n-butane
1 isobutane ⇌ 1 n-butane
Therefore, the equilibrium constant for the reverse reaction (n-butane to isobutane) can be calculated using the following equation:
K_reverse = 1 / K_forward
Substituting the given value:
K_reverse = 1 / 2.5
K_reverse = 0.4
Thus, the value of the equilibrium constant for the reverse reaction is 0.4.