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Topic: caculating value of k  (Read 5197 times)

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Offline newbie

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caculating value of k
« on: January 07, 2008, 04:04:30 PM »
Given K = 3.50 at 45°C for the reaction      A(g) + B(g) ↔ C(g)
and K = 7.10 at 45°C for the reaction         2A(g) + D(g) ↔ C(g)

a.   What is the value of K at the same temperature for the reaction    C(g) + D(g) ↔ 2B(g)

b.   What is the value of Kp at 45°C for the reaction?

c.   Starting with 1.50 atm partial pressures of both C and D, what is the mole fraction of B once equilibrium is reached?


I don't get how I would go about this problem. Can someone help me?

Offline Kryolith

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Re: caculating value of k
« Reply #1 on: January 07, 2008, 04:23:29 PM »
You have three equilibrium constants for the three reactions. I just call them K1, K2 and K3. What are the equations for them? Use the mass action law.

Offline newbie

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Re: caculating value of k
« Reply #2 on: January 13, 2008, 07:14:34 PM »
Yeah I did the equation for them but I still don't know how to find the value of each variable. This was recently on a test of mine but I didn't know how to do it.
Can you give me any more hints?

Offline Yggdrasil

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Re: caculating value of k
« Reply #3 on: January 13, 2008, 07:18:02 PM »
Kryolith's approach is one way to solve the problem.  A somewhat less efficient way of solving the problem (but perhaps a way that makes more intuitive sense) is to use the equilibrium constants to calculate a value of ΔG for each reaction.  Then combine (i.e. add or subtract) the two reactions in such a way that you can find the ΔG for the   C(g) + D(g) ↔ 2B(g) reaction.

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