For the reaction 2HI(.g.) leftharpoons H2(.g.)+I2(.g.) . The value of Kc is 4. If 2 mol of H2 , 2 mol of I2 and 2 mol of HI are present in one litre container then moles of I2 present at equilibrium is:

Question

For the reaction 2HI(.g.) leftharpoons H2(.g.)+I2(.g.) . The value of Kc is 4. If 2 mol of H2 , 2 mol of I2 and 2 mol of HI are present in one litre container then moles of I2 present at equilibrium is: (A) 0.8 (B) 3.2 (C) 2.4 (D) 4.4

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/c-dulbrvm5pst3.png" /> kc=4=((2 + x)2/(2 - 2 x)2) 2=(2 + x/2 - 2 x) X=0.4 Mol of I2 at equilibrium =2+0.4=2.4

Q. For the reaction $2 H I (g) ⇌ H_{2} (g) + I_{2} (g)$ . The value of $K_{c}$ is 4. If 2 mol of $H_{2}$ , 2 mol of $I_{2}$ and 2 mol of HI are present in one litre container then moles of $I_{2}$ present at equilibrium is:

0.8

3.2

2.4

4.4

$k_{c} = 4 = \frac{( 2 + x ) ^{2}}{( 2 - 2 x ) ^{2}}$

$2 = \frac{2 + x}{2 - 2 x}$

$X = 0.4$

Mol of $I_{2}$ at equilibrium $= 2 + 0.4 = 2.4$

Q. For the reaction 2HI(g)⇌H2​(g)+I2​(g) . The value of Kc​ is 4. If 2 mol of H2​ , 2 mol of I2​ and 2 mol of HI are present in one litre container then moles of I2​ present at equilibrium is:

0.8

3.2

2.4

4.4

kc​=4=(2−2x)2(2+x)2​ 2=2−2x2+x​ X=0.4 Mol of I2​ at equilibrium =2+0.4=2.4

Q. For the reaction $2 H I (g) ⇌ H_{2} (g) + I_{2} (g)$ . The value of $K_{c}$ is 4. If 2 mol of $H_{2}$ , 2 mol of $I_{2}$ and 2 mol of HI are present in one litre container then moles of $I_{2}$ present at equilibrium is:

$k_{c} = 4 = \frac{( 2 + x ) ^{2}}{( 2 - 2 x ) ^{2}}$

$2 = \frac{2 + x}{2 - 2 x}$

$X = 0.4$

Mol of $I_{2}$ at equilibrium $= 2 + 0.4 = 2.4$