5. a. The reaction below is endothermic with AH = +56.9 kJ mol". N0.(8) = 2NO,(8...

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Answer 1

Answer #1

Given deltaH = +Ve ; so it is endothermic reaction

N2O4 is a reactant. Adding reactant to equilibrium mixture will increase the formation of products.

Since NO2 is the product here, so equilibrium shift to right side or product side.

Reducing the volume at equilibrium shift the equilibrium to less moles side.

Here we have 1mole on left side and 2 moles on right side. So left side reaction is favored.

So NO2 formation will decreases.

Increasing Pressure at equilibrium shift the equilibrium to less moles side.

Here we have 1mole on left side and 2 moles on right side. So left side reaction is favored.

So NO2 formation will decreases.

Adding catalyst will increase the rate of both forward and back ward reactions equally.

So the net effect is null on equilibrium

Answer no change in the NO2 formation

Answer 2

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