Question

Consider the following gas-phase reaction: \[ 2SO_2 + O_2 \;\rightleftharpoons\; 2SO_3 \] If the enthalpy of reaction is negative, which condition promotes a higher equilibrium concentration of \(SO_3\)?

• A. Higher pressure and higher temperature
• B. Higher pressure and lower temperature
• C. Lower pressure and higher temperature
• D. Lower pressure and lower temperature

Hint:

For gas-phase equilibria: fewer gas moles favored by higher pressure; exothermic reactions favored by lower temperature. Always apply Le Chatelier's principle.

Answer 1

The Correct Option is B

Step 1: Effect of pressure (Le Chatelier's principle).
Left side: 3 moles of gas (2SO\(_2\) + O\(_2\)), Right side: 2 moles of gas (2SO\(_3\)). \(\Rightarrow\) Increasing pressure favors the forward reaction, since it reduces total moles of gas. Step 2: Effect of temperature.
The reaction is exothermic (\(\Delta H < 0\)). Lowering temperature favors the forward direction (formation of SO\(_3\)), since the system counteracts heat removal by producing more heat. Step 3: Check options.
- (A) High P, High T → High P favors forward, but High T opposes (endothermic shift).
- (B) High P, Low T → Both factors favor forward reaction (maximum SO\(_3\)).
- (C) Low P, High T → Both factors oppose forward reaction.
- (D) Low P, Low T → Low P opposes, though Low T favors. Step 4: Conclude.
Thus, the equilibrium concentration of SO\(_3\) is maximized at: \[ \boxed{\text{Higher pressure and lower temperature}} \]