The following reaction has been allowed to reach equilibrium at a temperature of

Question

The following reaction has been allowed to reach equilibrium at a temperature of 800 K: 2SO_2(g) + O_2 reversible 2SO_3(g) At this temperature, the equilibrium concentration of SO_2 was observed to be 3.0 middot 10^-3 M, that of O_2 was 3.5 middot 10^-3 M, and that of SO_3 was 5.0 middot 10^-2 M. The equilibrium constant, K_c, at 800 K, is: 4.4 middot 10^-8 1.3 middot 10^-5 2.1 middot 10^-4 9.0 middot 10^-2 7.1 middot 10^-1 4.8 middot 10^3 7.9 middot 10^4 2.3 middot 10^7 The reaction: 3Fe(s) + 4H_2 O(g) reversible Fe_3 O_4(s) + 4H_2(g) is allowed to reach equilibrium. Subsequently, some Fe_3 O_4 is removed from the reaction mixture. As a result, the equilibrium: Will shift to the right Will shift to the left Will not shift Cannot tell

Explanation / Answer

For the first question, we already have the equilibrium concentration, and according to the reaction:

2SO2 + O2 <--------> 2SO3  Kc = ?

Kc = [SO3]2 / [SO2]2[O2] ---> replacing the values of concentration we have:

Kc = (0.05)2 / (0.003)2(0.0035)
Kc = 7.94x104

For the second question, this is mostly theoric but whenever you substract a certain quantity of either the reactant or product, the equilibrium will shift to the side where the loss is to compensate that loss. In this case we are losing in the products side, so equilibrium will shift to the right (product).

Hope this helps

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