Hi! is my answer correct?
Consider the following chemical reaction:
H2 (g) + I2 (g) <> 2HI (g)
At equilibrium in a particular experiment, the concentrations of H2, I2, and HI were 0.15 M,
0.033 M, and 0.55 M, respectively. The value ofKeq for this reaction is
I solved by:
0.55 M^2 (bc of 2 on 2HI)
-----------------
(0.033 M)(0.15 M)
= 61
Looks good. Congrats. I would have put the squared sign over the 0.55 since it's the number you are squaring and not the unit. Technically, although molarity is what we substitute, Keq has no units. (Because the numbers that go there are activities. We use molarities in 99% of the problems you will see).
Did you know that chemists never die? They just stop reacting! *ba dum tss*
Now, let's take a look at your question. It seems like you've calculated the equilibrium constant (Keq) correctly. The value you obtained, 61, is indeed the correct Keq for this particular chemical reaction. Well done!
If you have any more chemistry questions, feel free to ask!
To calculate the value of Keq for the reaction, you need to write the expression for Keq using the concentrations of the species at equilibrium. Given the reaction:
H2 (g) + I2 (g) ⇌ 2HI (g)
The expression for Keq is:
Keq = [HI]^2 / ([H2] * [I2])
Substituting the given values:
Keq = (0.55 M)^2 / (0.033 M * 0.15 M)
Calculating this expression, we get:
Keq ≈ 603.03
So, the value of Keq for this reaction is approximately 603.03.
Yes, your answer is correct. You correctly applied the equation for calculating the equilibrium constant (Keq) using the concentration values of the reactants and products.
To find the value of Keq, you used the concentrations of the reactants and products at equilibrium. Since the coefficients for H2 and I2 are both 1, and the coefficient for HI is 2, you squared the concentration of HI (0.55 M) in the numerator.
In the denominator, you multiplied the concentrations of H2 (0.15 M) and I2 (0.033 M) as they appear in the balanced chemical equation.
By plugging in these values and performing the calculation, you obtained a Keq value of 61. Well done!