An equilibrium mixture of PCl 5 ( g ) , PCl 3 ( g ) , and Cl 2 ( g ) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respectively. A quantity of Cl 2 ( g ) is injected into the mixture, and the total pressure jumps to 263.0 Torr at the moment of mixing. The system then re-equilibrates. The chemical equation for this reaction is PCl 3 ( g ) + Cl 2 ( g ) − ⇀ ↽ − PCl 5 ( g ) Calculate the new partial pressures, P , after equilibrium is reestablished.

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

And the equilibrium constant at the reaction's temperature is:

Now, even when chlorine is added, such pressure equilibrium constant does not change, therefore, since the initial total pressure is:

The new pressures, due to the change owing to the chlorine's addition, turn out:

Therefore, the added chlorine is:

Thus, the new partial pressures are found via the law of mass action in terms of the change as follows:

Solving for one obtains:

Finally, the new partial pressures result:

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the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.

it forms an anion that has a larger radius. anions have a negative charge and are larger than the original atom.