At a certain temperature, 1.0 mole of PCl3(g) and 2.0 mole of Cl2(g) were placed in a 3.0 litre container. When equilibrium is reached, only 0.70 mole of PCl3 remained unreacted. Calculate the value of Kc for the reaction: PCl3(g) +Cl2(g) leftharpoons PCl5(g)

Question

At a certain temperature, 1.0 mole of PCl3(g) and 2.0 mole of Cl2(g) were placed in a 3.0 litre container. When equilibrium is reached, only 0.70 mole of PCl3 remained unreacted. Calculate the value of Kc for the reaction: PCl3(g) +Cl2(g) leftharpoons PCl5(g) (A) 0.76 (B) 0.076 (C) 0.38 (D) 1.52

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<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/c1cfa93afa34ab9d770de74104e4f053-.png" /> Kc=([PCl5]/[PCl3][Cl2])=(0.3 / 3/(0.7)3×(1.7)3=(0.9/0.7×1.7)=0.76

$0.76$

$0.076$

$0.38$

$1.52$

$K_{c} = \frac{[ PC l _{5} ]}{[ PC l _{3} ] [ C l _{2} ]} = \frac{0.3/3}{\frac{0.7}{3} \times \frac{1.7}{3}} = \frac{0.9}{0.7 \times 1.7} = 0.76$

Q. At a certain temperature, 1.0mole of PCl3(g)​ and 2.0mole of Cl2(g)​ were placed in a 3.0litre container. When equilibrium is reached, only 0.70mole of PCl3​ remained unreacted. Calculate the value of Kc​ for the reaction : PCl3(g)​+Cl2(g)​⇌PCl5(g)​

0.76

0.076

0.38

1.52

Kc​=[PCl3​][Cl2​][PCl5​]​=30.7​×31.7​0.3/3​=0.7×1.70.9​=0.76

$0.76$

$0.076$

$0.38$

$1.52$

$K_{c} = \frac{[ PC l _{5} ]}{[ PC l _{3} ] [ C l _{2} ]} = \frac{0.3/3}{\frac{0.7}{3} \times \frac{1.7}{3}} = \frac{0.9}{0.7 \times 1.7} = 0.76$