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Chemistry · EQUILIBRIUM
हिन्दीInitially, \( 3.00\text{ mol} \) of \( \text{PCl}_5 \) is kept in a \( 1\text{ L} \) closed reaction vessel at \( 380\text{ K} \). It dissociates as \( \text{PCl}_5(\text{g}) \rightleftharpoons \text{PCl}_3(\text{g}) + \text{Cl}_2(\text{g}) \) with \( K_c = 1.80 \). At equilibrium, what is the concentration of \( \text{PCl}_5 \)?
Correct answer · Option 2
\( 1.41\text{ M} \)
Explanation
\( K_c = \frac{x^2}{3-x} = 1.80 \implies x^2 + 1.8x - 5.4 = 0 \implies x = 1.59\text{ M} \). Therefore, \( [\text{PCl}_5] = 3.00 - 1.59 = 1.41\text{ M} \).
