To calculate the temperature at which a reaction becomes spontaneous, we need to use Gibbs free energy ($\Delta G$) and the relationship between enthalpy ($\Delta H$), entropy ($\Delta S$), and temperature ($T$):

$\Delta G = \Delta H - T \Delta S$

For a reaction to be spontaneous, $\Delta G$ must be negative. Therefore, the reaction becomes spontaneous when:

$\Delta G = 0$

This gives us the condition:

$T = \frac{\Delta H}{\Delta S}$

Where:

• $\Delta H$ is the change in enthalpy of the reaction (usually given in $\text{kJ/mol}$),
• $\Delta S$ is the change in entropy of the reaction (usually given in $\text{J/mol·K}$).

Steps:

1. Find the values for $\Delta H$ and $\Delta S$ for the reaction: $\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$ These values are typically available in thermodynamic tables.

2. Convert $\Delta H$ to $\text{J/mol}$ if it's given in $\text{kJ/mol}$ (since $\Delta S$ is usually in $\text{J/mol·K}$).

3. Apply the equation $T = \frac{\Delta H}{\Delta S}$.

Once you have the values, you can plug them into the formula to determine the temperature at which the reaction becomes spontaneous.

Do you have the values for $\Delta H$ and $\Delta S$? If not, I can help explain how to look them up.

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Would you like a detailed explanation of the next steps?

Here are 5 related questions to expand the discussion:

1. What role does entropy play in determining spontaneity?
2. How does the sign of $\Delta H$ affect the spontaneity of a reaction?
3. What is the significance of the equilibrium constant in relation to Gibbs free energy?
4. How would the spontaneity change if $\Delta S$ were negative instead of positive?
5. Can a reaction be non-spontaneous at one temperature and spontaneous at another?

Tip: A reaction with positive $\Delta H$ and positive $\Delta S$ tends to be spontaneous at higher temperatures!