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You and a friend are studying for a chemistry exam. What if your friend says, “Adding an inert gas to a system of gaseous components at equilibrium never changes the equilibrium position”? How do you explain to your friend that this holds true for a system at constant volume but is not necessarily true for a system at constant pressure? When would it hold true for a system at constant pressure?

Short Answer

Expert verified

You need to explain to your friend that addition of inert gas does not change the equilibrium only at constant volume but not at constant pressure.

Step by step solution

01

Addition of inert gas at constant volume

Addition of inert gas increases the total pressure at constant volume. But inert gas has no effect on concentrations and partial pressures of reactants and products. Hence, inert gas also has no effect on equilibrium constant. Therefore, addition of an inert gas to a system of gaseous components at equilibrium does not change the equilibrium position at constant volume condition.

02

Addition of inert gas at constant pressure

Addition of inert gas to the system at constant pressure increases the volume. Therefore, the no of moles per unit volume decreases.To nullify this condition, the equilibrium re-established itself to such a direction where the no of moles is higher. Therefore, addition of an inert gas to a system of gaseous components at equilibrium changes the equilibrium position at constant pressure condition.

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Most popular questions from this chapter

Question: For the reaction

H2(g)+Br2(g)2HBr(g)

Kp = 3.5×104 at 1495 K. What is the value of Kp for the following reactions at 1495 K?

a. role="math" localid="1649230142685" HBr(g)12H2(g)+12Br2(g)

b. 2HBr(g)H2(g)+Br2(g)

c. 12H2(g)+12Br2(g)HBr(g)

The compound SbCl5(g) decomposes at high temperatures to gaseous antimony trichloride and chlorine gas. When 89.7 g of SbCl5(g) is placed in a 15.0-L container at 1800C, the SbCl5(g) is 29.2% decomposed (by moles) after the system has reached equilibrium.

a. Calculate the value of K for this reaction at 1800C.

b. Determine the number of moles of chlorine gas that must be injected into the flask to make the new equilibrium pressure of antimony trichloride half that of the original equilibrium pressure of antimony trichloride in the original experiment.

Consider the reaction

P4(g)2P2(g)

where Kp= 1.00 × 10-1 at 1325 K. In an experiment where P4(g) was placed in a container at 1325 K, the equilibrium mixture of P4(g) and P2(g) has a total pressure of 1.00 atm. Calculate the equilibrium pressures of P4(g) and P2(g). Calculate the fraction (by moles) of P4(g) that has dissociated to reach equilibrium.

Suppose 1.50 atm of CH4(g), 2.50 atm of C2H6(g), and 15.00 atm of O2(g) are placed in a flask at a given temperature. The reactions are

CH4+2O2CO2+2H2OKp=1.00×1042C2H6+7O24CO2+6H2OKp=1.00×108

Calculate the equilibrium pressures of all gases.

The gas arsine (AsH3) decomposes as follows:

2AsH3(g) 34 2As(s) 1 3H2(g)

In an experiment pure AsH3(g) was placed in an empty,rigid, sealed flask at a pressure of 392.0 torr. After 48 hthe pressure in the flask was observed to be constant at488.0 torr.

a. Calculate the equilibrium pressure of H2(g).

b. Calculate Kp for this reaction.

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