Question

The value of equilibrium constant for the two equilibriums at 300 K are: (i) A + B 2C K = 4 (ii) P + Q 2R K = 16 Which of the two given equilibrium has the greater concentration of product if we start with one mole of each of A, B , P and Q? Explain​

Answer 1

Answer:

Explanation:

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Answer 2

The concentration of product would be greater in reaction 2.

Explanation:

Reaction 1:                     $A + B\rightleftharpoons 2C; K_c =4$

Initial Concentration      1      1        0

Change                           -x    -x     +2x

At equilibrium               (1-x)   (1-x)   2x

Equilibrium constant $K_c = \frac{[C]^2}{[A][B]}$

Substituting the values

$4 = \frac{(2x)^2}{(1-x)(1-x)}$

$4(1-x)^2 = 4x^2$

$2(1-x) = 2x$

$2-2x = 2x$

$2= 2x+2x$

$2 = 4x$

$x = 0.5$

Therefore product concentration, [C] = 2x $= 2 \times 0.5 = 1 M$

Similarly, for reaction 2:

Reaction 2:                      $P + Q\rightleftharpoons 2R; K_c =16$

Initial Concentration      1      1        0

Change                           -x    -x     +2x

At equilibrium               (1-x)   (1-x)   2x

Equilibrium constant $K_c = \frac{[R]^2}{[P][Q]}$

Substituting the values

$16 = \frac{(2x)^2}{(1-x)(1-x)}$

$16(1-x)^2 = 4x^2$

$4(1-x) = 2x$

$4-4x = 2x$

$4= 2x+4x$

$4 = 6x$

$x = 4/6 = 0.66$

Therefore product concentration, [R] = 2x $= 2 \times 0.66 = 1.32 M$

Hence, concentration of product would be greater in reaction 2.