Question

4 moles of A are mixed with 4 moles o B When 2 moles of C are formed at equilibrium according to the reaction A+B=C+D. The equilibrium constant is? ​

Answer 1

Explanation:

A + B ⇄ C + D

initial mole - 4 4 0 0

at equilibrium - 4-x 4-x x=2 x=2

so final moles - 2 2 2 2

the equilibrium constant = [C][D] / [A][B] = [2][2] / [2][2] = 1

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Answer 2

The equilibrium constant of the reaction is 1

Explanation:

We are given:

Initial moles of A = 4 moles

Initial moles of B = 4 moles

Equilibrium moles of C = 2 moles

For the given chemical equation:

                  $A+B\rightleftharpoons C+D$

Initial:             4     4

At eqllm:       4-x   4-x        x    x

Calculating the value of 'x':

$\Rightarrow x=2$

So, equilibrium concentration of A = $(4-x)=4-2=2$

Equilibrium concentration of B = $(4-x)=4-2=2$

Equilibrium concentration of D = $x=2$

The expression of $K_{eq}$ for above equation follows:

$K_{eq}=\frac{[C][D]}{[A][B]}$

Putting values in above equation, we get:

$K_{eq}=\frac{2\times 2}{2\times 2}\\\\K_{eq}=1$

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