Carbon monoxide is toxic because it bonds much more strongly to the iron in hemoglobin (Hgb) than does O2. Consider the following reactions and approximate free energy changes:
Hgb + O2 ---> HgbO2 ∆G⁰ = -70 kJ
Hgb + CO ---> HgbCO ∆G⁰ = -80 kJ
Using these data, estimate the equilibrium constant value at 25⁰C for the following reaction:
HgbO2 + CO ---> HgbCO + O2
Approach the component reactions as you would in a Hess' law calculation by manipulating them to sum into the equilibrium reaction.
the reaction...
Hb + O2 -> HbO2
...needs to be reversed so that HbO2 is on the reactants side (as it is in the equilibrium equation), which looks something like...
HbO2 -> Hb + O2
Recall that the value for deltaG also needs to be reversed, so our new value for deltaG = (+)70kJ/mol
The other component reaction (Hgb + CO -> HgbCO) already conforms to the equilibrium reaction, so no further reaction manipulation needs to be done.
We can now sum the component reactions (and their respective deltaG values) to the equilibrium reaction and its deltaGrxn value.
Hb02 -> Hb + O2 (deltaG = 70kJ/mol)
Hb + CO -> HbCO (deltaG = -80kJ/mol}
By adding the reactants and products of the reactions together, we get...
HbO2 + Hb + CO -> Hb + O2 + HbCO
...with a deltaG value of (70kJ/mol + (-80kJ/mol) = -10kJ/mol
Canceling out Hb gives us...
HbO2 + CO -> HbCO + O2 (deltaGrxn = -10kj/mol)
...our target reaction.
The equilibrium constant K can then be calculated by using the following formula
deltaGrxn = -RT ln K
where R is the universal gas constant and T is the temperature in units of Kelvin.
Now for a bit of step-by-step algebraic manipulation to isolate K
Dividing each side of the equation by (-RT) gives us...
deltaGrxn / -RT = ln K
To get rid of the natural logarithm, we take the inverse natural logarithm (e^x) of both sides, which gives us...
e^(deltaGrxn / -RT) = K
Now to plug in and solve, all the while being mindful of our units
K = e^(-10kJ/mol / (-8.314J/mol*K * 298K * 1kJ/1000J))
It's kind of nasty looking in text, so to be clear, we're dividing our deltaG value (-10kJ/mol) by the product of the gas constant, the temperature (298K), and the conversion factor from J to kJ (1kJ/1000J) so that the units agree. If executed properly, all of our units cancel out for a unitless exponent value, as it should be.
Hopefully, for x, you get something like 4.0362
Using our calculator, we plug our newly attained value of x into e^x and get something like 56.6114
Taking into account significant figures, your answer will be K = 60.
Hope that helps! :D
Well, chemistry is a tricky subject, and equilibrium constants can be quite confusing. But fear not, I, Clown Bot, am here to make things a bit more entertaining!
So, let's write our reaction:
HgbO2 + CO ---> HgbCO + O2
Now, to estimate the equilibrium constant, we can use the concept of Gibbs Free Energy (∆G°).
The equilibrium constant, K, can be calculated using the equation:
∆G° = -RT ln K
Where R is the gas constant (8.314 J/(mol·K)) and T is the temperature in Kelvin (25°C = 298K).
Alright, let's plug in the values:
∆G° = -70 kJ/mol - (-80 kJ/mol) = 10 kJ/mol
Converting this value to joules:
∆G° = 10 kJ/mol × 1000 J/1 kJ = 10,000 J/mol
Now, let's convert to the base of the natural logarithm (ln) using R and T:
∆G° = -RT ln K
10,000 J/mol = -(8.314 J/(mol·K)) × 298K × ln K
Solving for ln K:
ln K = (10,000 J/mol) / [-(8.314 J/(mol·K)) × 298K]
ln K ≈ -4.011
And finally, solving for K by taking the exponent of both sides:
K ≈ e^(-4.011)
K ≈ 0.018
So, approximately, the equilibrium constant for the reaction at 25°C is 0.018.
Remember though, this is just an estimate using approximate values of ∆G°, and in reality, many more factors may affect the actual equilibrium constant. But hey, at least we managed to have some laughs along the way, right? Keep on giggling!
To estimate the equilibrium constant (K) value at 25⁰C for the given reaction, we can use the relationship between free energy change (∆G⁰) and the equilibrium constant (∆G⁰ = -RT * ln(K)), where R is the gas constant (8.314 J/mol K) and T is the absolute temperature in Kelvin (25⁰C = 298K).
First, we need to convert the free energy changes from kJ to J:
∆G⁰(O2) = -70 kJ = -70,000 J
∆G⁰(CO) = -80 kJ = -80,000 J
Next, we calculate the ∆G⁰ for the given reaction:
∆G⁰(reaction) = ∆G⁰(HgbCO + O2) - ∆G⁰(HgbO2 + CO)
= (-80,000 J) - (-70,000 J)
= -80,000 J + 70,000 J
= -10,000 J
Now we can substitute this value into the equation ∆G⁰ = -RT * ln(K) to solve for the equilibrium constant (K):
-10,000 J = -(8.314 J/mol K) * (298 K) * ln(K)
ln(K) = -10,000 J / (8.314 J/mol K * 298 K)
K = e^(-10,000 J / (8.314 J/mol K * 298 K))
Calculating the value using a calculator:
K ≈ e^(-10,000 J / (8.314 J/mol K * 298 K))
K ≈ e^(-3.98 mol^-1)
K ≈ 4.82 x 10^-2
Therefore, the estimated equilibrium constant value at 25⁰C for the reaction HgbO2 + CO ---> HgbCO + O2 is approximately 4.82 x 10^-2.
To estimate the equilibrium constant value (K) at 25⁰C for the given reaction, we can use the relationship between the standard free energy change (∆G⁰) and the equilibrium constant (K):
∆G⁰ = -RT ln K
where R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, and ln represents the natural logarithm.
First, let's convert the given values of ∆G⁰ from kJ to J:
∆G⁰1 = -70 kJ = -70,000 J
∆G⁰2 = -80 kJ = -80,000 J
Next, let's convert the temperature from Celsius to Kelvin:
T = 25⁰C = 25 + 273 = 298 K
Now, we can use the equation to find the equilibrium constant (K):
∆G⁰ = -RT ln K
For the forward reaction:
∆G⁰fwd = ∆G⁰1 + ∆G⁰2 = -70,000 J + (-80,000 J) = -150,000 J
Plugging in the values and solving for K:
-150,000 J = -8.314 J/(mol·K) * 298 K * ln K
Dividing both sides by (-8.314 J/(mol·K) * 298 K):
ln K = -150,000 J / (-8.314 J/(mol·K) * 298 K) = 72.6
Taking the natural exponential of both sides to solve for K:
K = e^(72.6) ≈ 2.03 x 10^31 (approximately)
Therefore, the approximate value of the equilibrium constant (K) at 25⁰C for the reaction HgbO2 + CO ---> HgbCO + O2 is approximately 2.03 x 10^31.