At 700 k the equilibrium constant Kc for the reaction between NO(g) and O2(g) forming NO2(g) is 8.7*10^6. The rate constant for the reverse reaction at this temperature is 0.54 M^(-1)s^(-1). What is the value of the rate constant for the forward reaction at 700 K?
Why did the reaction go on a diet?
Because it wanted to find the rate constant!
But in all seriousness, to find the rate constant for the forward reaction at 700 K, you can use the equilibrium constant (Kc) and the rate constant for the reverse reaction.
For any reversible reaction, the equilibrium constant (Kc) is equal to the ratio of the rate constant for the forward reaction (kf) to the rate constant for the reverse reaction (kr).
So, we can set up the following equation:
Kc = kf / kr
Plugging in the given values:
8.7 * 10^6 = kf / 0.54
Now, to find kf, we rearrange the equation:
kf = Kc * kr
kf = (8.7 * 10^6) * (0.54)
kf = 4.7 * 10^6 M^(-1)s^(-1)
So, the value of the rate constant for the forward reaction at 700 K is approximately 4.7 * 10^6 M^(-1)s^(-1). Happy reacting!
To find the value of the rate constant for the forward reaction at 700 K, we can use the relationship between the equilibrium constant (Kc) and the rate constants (k forward and k reverse).
The equilibrium constant (Kc) is defined as the ratio of the rate constants for the forward (k forward) and reverse (k reverse) reactions:
Kc = k forward / k reverse
Given that Kc = 8.7 * 10^6 and k reverse = 0.54 M^(-1)s^(-1), we can rearrange the equation and solve for k forward:
k forward = Kc * k reverse
Substituting the given values:
k forward = (8.7 * 10^6) * (0.54 M^(-1)s^(-1))
k forward ≈ 4.698 M^(-1)s^(-1)
Therefore, the value of the rate constant for the forward reaction at 700 K is approximately 4.698 M^(-1)s^(-1).
To find the value of the rate constant (k_forward) for the forward reaction at 700 K, you can use the equilibrium constant expression and the rate constant for the reverse reaction.
The equilibrium constant expression for the reaction is given as:
Kc = [NO2] / ([NO] * [O2])
where [NO2], [NO], and [O2] represent the molar concentrations of NO2, NO, and O2, respectively.
Given that Kc = 8.7 * 10^6 and the concentrations at equilibrium are in the units of moles per liter (M), we can express the equilibrium concentrations as:
[NO2] = 700k
[NO] = x (unknown)
[O2] = x (unknown)
Since we don't know the initial concentrations of NO and O2, we can represent them as 'x'.
Now, we can rearrange the equilibrium constant expression to solve for [NO]:
[NO] = [NO2] / (Kc * [O2])
= (700k) / (8.7 * 10^6 * x)
= (700 / 8.7) * (k / x) (Equation 1)
The rate constant for the reverse reaction, k_reverse, is given as 0.54 M^(-1)s^(-1). At equilibrium, the rates of the forward and reverse reactions are equal. Therefore, we can equate the rate constants:
k_forward * [NO]^1 * [O2]^1 = k_reverse * [NO2]^1
k_forward * [NO] * [O2] = k_reverse * [NO2]
k_forward * x * x = 0.54 * (700k)
Substituting the value of [NO] from Equation 1 and simplifying:
(k / x) * x * x = 0.54 * (700k)
x^2 = (0.54 * 700) * x
x^2 - 378x = 0
Solving this quadratic equation for x, you'll find two possible values of x. However, we know that the molar concentration cannot be negative, so we can discard the negative value.
Hence, the value of [NO] at equilibrium is x = 378. Now, we can substitute this value back into Equation 1 to find the rate constant for the forward reaction:
k_forward = (700 / 8.7) * (k / x)
= (700 / 8.7) * (k / 378)
= 109.521 * (k / 378)
= 0.2893 * k
Therefore, the value of the rate constant (k_forward) for the forward reaction at 700 K is approximately 0.2893 * k, where k represents the rate constant for the reverse reaction.