Initially 3 moles of ' A ' was taken in a text1 textL container. The approx. moles of A left in the container when the following equilibrium established is x× 10- 6 .The value of ' x ' is. 3A leftharpoons B;KC=8× 1015

Question

Initially 3 moles of ' A ' was taken in a text1 textL container. The approx. moles of A left in the container when the following equilibrium established is x× 10- 6 .The value of ' x ' is. 3A leftharpoons B;KC=8× 1015 (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/c-y4cvrrme7rml.png" /> Now, (1 - x/(3 x 3)=8× 1015 3x=5× 10- 6

Q. Initially 3 moles of ' $A$ ' was taken in a $1 L$ container. The approx. moles of A left in the container when the following equilibrium established is $x \times 1 0^{- 6}$ .The value of ' $x$ ' is.

$3 A ⇌ B; K_{C} = 8 \times 1 0^{15}$

Now,

$\frac{1 - x}{( 3 x ^{3}} = 8 \times 1 0^{15}$

$3 x = 5 \times 1 0^{- 6}$

Q. Initially 3 moles of ' A ' was taken in a 1L container. The approx. moles of A left in the container when the following equilibrium established is x×10−6 .The value of ' x ' is. 3A⇌B;KC​=8×1015

Now, (3x31−x​=8×1015 3x=5×10−6

Q. Initially 3 moles of ' $A$ ' was taken in a $1 L$ container. The approx. moles of A left in the container when the following equilibrium established is $x \times 1 0^{- 6}$ .The value of ' $x$ ' is.

$3 A ⇌ B; K_{C} = 8 \times 1 0^{15}$

Now,

$\frac{1 - x}{( 3 x ^{3}} = 8 \times 1 0^{15}$

$3 x = 5 \times 1 0^{- 6}$