Consider this reaction at equilibrium 2SO2(g) + O2(g) 2SO3(g), delta H degree rx

Question

Consider this reaction at equilibrium 2SO2(g) + O2(g) 2SO3(g), delta H degree rxn = -198 kJ/mol It the volume of the system is compressed at constant temperature, what change will occur in the position of the equilibrium? a stift to produce more SO2 a shift to produce more O2 no change a shift to produce more SO3 Hydrogen iodide decomposes according to the equation 2HI(g) H2(g) + I2(g) for which Kc = 0.0156 at 400 degree C. 0.550 mol HI was injected into a 2.00 L reaction vessel at 400 degree C. Calculate the concentration of H2 at equilibrium.

Explanation / Answer

2SO2+O2 <--> 2SO3 H = -198

If you compress the system, you will favour the side with less moles, that is, the produc side (has only 2 moles wherease the reactants have 3 moles).

The answer is D) the shift produces more SO3

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