For the system A-g-2B-g- rightleftharpoons C-g- the equilibrium concentrations are A-0-06 mol L-1-B-0-12 mol L-1-C-0-216 mol L-1-Then K-c- the reaction is -250450525416

Question

Toppr Admin · Accepted Answer

For Reaction A-2B-x21CC-C -xA0-Kc-C-B-2-A-By Substituting the values- We getKc - 250

For the system $A_{(g)} + 2 B_{(} g) ⇌ C_{(g)}$ , the equilibrium concentrations are
$A = 0.06 m o l L^{- 1}$
$B = 0.12 m o l L^{- 1}$
$C = 0.216 m o l L^{- 1}$
Then $K_{c}$ for the reaction is _____.
$250$
$450$
$525$
$416$

For Reaction $A + 2 B ⇌ C$ ,

$K_{c} = \frac{[C]}{{[B]}^{2} [A]}$

By Substituting the values, We get

$K_{c}$ = 250

For the system A(g)+2B(g)⇌C(g), the equilibrium concentrations are A=0.06 mol L−1B=0.12 mol L−1C=0.216 mol L−1Then Kc for the reaction is _____.250450525416

For Reaction A+2B⇌C , Kc=[C][B]2[A]By Substituting the values, We getKc = 250

For the system $A_{(g)} + 2 B_{(} g) ⇌ C_{(g)}$ , the equilibrium concentrations are
$A = 0.06 m o l L^{- 1}$
$B = 0.12 m o l L^{- 1}$
$C = 0.216 m o l L^{- 1}$
Then $K_{c}$ for the reaction is _____.
$250$
$450$
$525$
$416$

For Reaction $A + 2 B ⇌ C$ ,

$K_{c} = \frac{[C]}{{[B]}^{2} [A]}$

By Substituting the values, We get

$K_{c}$ = 250