Two moles of HI were heated in a sealed tube at 440° C till the equilibrium was reached. HI was found to be 22 % decomposed. The equilibrium constant for dissociation is

Question

Two moles of HI were heated in a sealed tube at 440° C till the equilibrium was reached. HI was found to be 22 % decomposed. The equilibrium constant for dissociation is (A) 0.282 (B) 0.0796 (C) 0.0199 (D) 1.99

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<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/2581e1336104966e6ca66c0a0af826ea-.png" /> K=([ H 2][ I 2]/[ HI ]2)=(0.22 × 0.22/(1.56)2)=0.0199

Q. Two moles of $H I$ were heated in a sealed tube at $44 0^{\circ} C$ till the equilibrium was reached. HI was found to be $22%$ decomposed. The equilibrium constant for dissociation is

0.282

0.0796

0.0199

1.99

$K = \frac{[ H _{2} ] [ I _{2} ]}{[ H I ] ^{2}} = \frac{0.22 \times 0.22}{( 1.56 ) ^{2}} = 0.0199$

Q. Two moles of HI were heated in a sealed tube at 440∘C till the equilibrium was reached. HI was found to be 22% decomposed. The equilibrium constant for dissociation is