What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?

2ICl(g) ⇌ I2(g)+Cl2(g) ;  K=0.14

Asked by Pragya Singh | 11 months ago |  68

##### Solution :-

The given reaction is:

2ICl(g)   ⇌  I2(g)   +   Cl2(g)

Initial conc. 0.78 M,  0,   0

At equilibrium  (0.78-2x) M,    x M,    x M

Now, we can write,

$$\dfrac{[I_{2}][Cl_{2}]}{[IC]^{2}}=K_{c}\\ \\ \Rightarrow \dfrac{x\times x}{(0.78-2x)^{2}}=0.14\\ \\$$

$$\Rightarrow \dfrac{x^{2}}{(0.78-2x)^{2}}=0.14\\ \\$$

$$\Rightarrow \dfrac{x}{0.78-2x}=0.374\\ \\$$

$$\Rightarrow x=0.292-0.748x\\ \\$$

$$\Rightarrow 1.748x=0.292\\ \\$$

$$\Rightarrow x=0.167$$

Hence, at equilibrium,

$$[H_2]=[I_2]=0.167 M$$

$$[HI]= (0.78-2\times 0.167)M\\ \\$$

$$=0.446M$$

Answered by Abhisek | 11 months ago

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