The value of the equilibrium constant (Kp) as represented by the first chemical equation is 1.02 x 102 at 727 °C. Calculate the value of the equilibrium constant (Kp) for the second equation at the same temperature. Express answer in scientific notation.
2F(g) = F2(g)
F2(g) = 2F(g)
I'm not getting the math here, like how to mortify the first equation to equal it to the second one. [F]/[F]^2 = [F]^2/[F] ?
The answer is 9.8 x 10^-3
Wait never mind I figured it out.
1/F = F so 1/102 right?
right.
1/k = 1/1.02 x 10^2 = 9.8 x 10^-3
To calculate the value of the equilibrium constant (Kp) for the second equation at the same temperature, you need to use the concept of partial pressures and the stoichiometry of the reaction.
The equilibrium constant (Kp) is defined as the ratio of the partial pressures of the products to the partial pressures of the reactants, with each partial pressure raised to the power of its coefficient in the balanced equation.
Let's start with the first equation:
2F(g) ⇌ F2(g)
The equilibrium constant (Kp) for this equation is given as 1.02 x 10^2.
Now, let's consider the second equation:
F2(g) ⇌ 2F(g)
To determine the equilibrium constant (Kp) for this equation, we need to relate it to the first equation. We can invert the first equation to find the equation for the reverse reaction. Since the equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction, we have:
F2(g) ⇌ 2F(g)
(Kp)reverse = 1/(Kp)forward = 1/(1.02 x 10^2)
Now, we compare this equation to the second equation:
F2(g) ⇌ 2F(g)
(Kp)reverse = (Kp)second equation
Therefore, the equilibrium constant (Kp) for the second equation at the same temperature is 1/(1.02 x 10^2).
To express this answer in scientific notation, we need to divide 1 by 1.02 and adjust the exponent of 10. So, the value of the equilibrium constant (Kp) for the second equation is approximately 9.8 x 10^-3.