The reaction 2 AsH31g2 ∆ As21g2 + 3 H21g2 has Kp =
7.2 * 107 at 1073 K. At the same temperature, what is Kp
for each of the following reactions?
(c) 9 H21g2 + 3 As21g2 ∆ 6 AsH31g2

Question

The reaction 2 AsH31g2 ∆ As21g2 + 3 H21g2 has Kp =

7.2 * 107 at 1073 K. At the same temperature, what is Kp

for each of the following reactions?

(c) 9 H21g2 + 3 As21g2 ∆ 6 AsH31g2

Accepted Answer

Welcome back everyone. The reaction three moles of CO two in equilibrium with three moles of C and three moles of 02 has a KP value of 5.6 multiplied by 10 of the power of eight at 1200 Kelvin. What is the KP of the reaction below at the same temperature? So let's compare our reactions. The first one is 3 CO2 in equilibrium with three C N 302. The equilibrium constant for this reaction is given to us, let's call it KP one. Now we want to obtain an equilibrium constant of a different reaction which is nine C plus 9 +02 in equilibrium width nine co two. Can we notice the relationship between these two reactions? Well, essentially they have the same species. And also we can notice that the products of reaction one are actually the reactants of reaction number two. So first of all, we are flipping or reversing reaction one. After we flip it, we are also multiplying all of our coefficients by three, right, we can clearly see that three co two becomes 9 CO2, 3 C becomes nine C and 302 becomes 902. So the second transformation that we are performing is multiplying the coefficients by three. Now let's recall that whenever we are flipping an equilibrium reaction, the new equilibrium constant is simply the reciprocal of the original equilibrium constant. So our first step would result in a one divided by KP one. And finally, our second step tells us that we are multiplying our coefficients by three. So we're taking the previously calculated equilibrium constant and we want to raise it to the multiplier which is three. Let's recall that whenever we are multiplying a reaction by a specific coefficient to get the new equilibrium constant, we are simply raising our equilibrium constant to that multiplier or coefficient. In this case, it is equal to three. So our KP two, the equilibrium constant for the target equation is equal to the reciprocal of KP one raised to the power of three. So we are reversing it and multiplying all of our coefficients by three. Now, all that is loved is simply calculate the result. So we're taking the KP one value of 5.6 multiplied by 10 to the power of a, we're getting the reciprocal of it and raising the power of three, which gives us 5.7 multiplied by 10, the power of negative 27th, which is our final answer. Thank you for watching.

The reaction 2 AsH31g2 ∆ As21g2 + 3 H21g2 has Kp =
7.2 * 107 at 1073 K. At the same temperature, what is Kp
for each of the following reactions?
(c) 9 H21g2 + 3 As21g2 ∆ 6 AsH31g2

Verified Solution

Key Concepts

Equilibrium Constant (Kp)

Reaction Stoichiometry

Reverse Reaction and Kp Relationship

The reaction 2 AsH31g2 ∆ As21g2 + 3 H21g2 has Kp =7.2 * 107 at 1073 K. At the same temperature, what is Kpfor each of the following reactions?(c) 9 H21g2 + 3 As21g2 ∆ 6 AsH31g2

Verified Solution

Key Concepts

Equilibrium Constant (Kp)

Reaction Stoichiometry

Reverse Reaction and Kp Relationship

The reaction 2 AsH31g2 ∆ As21g2 + 3 H21g2 has Kp =
7.2 * 107 at 1073 K. At the same temperature, what is Kp
for each of the following reactions?
(c) 9 H21g2 + 3 As21g2 ∆ 6 AsH31g2