When S in the form of is heated at 900 K, the initial pressure of 1 atm falls by 30% at equilibrium. This is because of conversion of some (g) to (g). The of the reaction is:

Question

When S in the form of $S_8$ is heated at 900 K, the initial pressure of 1 atm falls by 30% at equilibrium. This is because of conversion of some $S_8$(g) to $S_2$(g). The $K_p$ of the reaction is:

Prashant Kotian · Accepted Answer

Correct answer: B. Step 1 - Write down the balanced reaction and identify initial conditions The thermal dissociation of sulfur vapor at $900\text{ K}$ is: $$\ce{S8(g) <=> 4S2(g)}$$ Initially, only \ce{S8(g)} is present at $1\text{ atm}$: $$p_{\ce{S8}, 0} = 1\text{ atm}, \quad p_{\ce{S2}, 0} = 0\text{ atm}$$ Step 2 - Construct the ICE Table in terms of partial pressures The partial pressure of \ce{S8} decreases by 30% of 1 atm = 0.30 atm. Let $x = 0.30\text{ atm}$. $$ \begin{array}{lccccc} \text{Species} & \ce{S8(g)} & \ce{<=>} & \ce{4S2(g)} \ \hline \text{Initial (atm)} & 1 & & 0 \ \text{Change (atm)} & -x & & +4x \ \text{Equilibrium (atm)} & 1 - x & & 4x \ \hline \end{array} $$ Substituting $x = 0.30\text{ atm}$: $$p_{\ce{S8}} = 1 - 0.30 = 0.70\text{ atm}$$ $$p_{\ce{S2}} = 4 \times 0.30 = 1.20\text{ atm}$$ Step 3 - Formulate the $K_p$ expression and calculate $$K_p = \frac{(p_{\ce{S2}})^4}{p_{\ce{S8}}} = \frac{(1.20)^4}{0.70} = \frac{2.0736}{0.70} \approx \boxed{2.96\text{ atm}^3}$$ Step 4 - Evaluate all options - **Option (A) 0.011 atm^-3**: Incorrect. Wrong sign on exponent — this would correspond to the reverse reaction. - **Option (B) 2.96 atm^3**: Correct. Derived from (1.20)^4 / 0.70 = 2.96 atm^3. - **Option (C) 1.71 atm^3**: Incorrect. Arises from incorrect stoichiometry (using coefficient 2 instead of 4 for S2). - **Option (D) 204.8 atm^3**: Incorrect. Calculation error in partial pressures.

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