10 moles of A2 , 15 moles of B2 and 5 moles of AB are placed in a 2 litre vessel and allowed the come to equilibrium. The final concentration of AB is 10.5 M , mathrmA2( mathrm~g)+ mathrmB2( mathrm~g) leftharpoons 2 mathrmAB( mathrmg) Determine the value of equilibrium constant (KC) for the reaction.

Question

10 moles of A2 , 15 moles of B2 and 5 moles of AB are placed in a 2 litre vessel and allowed the come to equilibrium. The final concentration of AB is 10.5 M , mathrmA2( mathrm~g)+ mathrmB2( mathrm~g) leftharpoons 2 mathrmAB( mathrmg) Determine the value of equilibrium constant (KC) for the reaction. (A) 25.3 (B) 31.5  (C) 36.3 (D) 40.5

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/c-xoowq1owavfr.png" /> (5 + 2 x/2)=10.5 ; textx = 8 KC=(([. A B ].)2/[A2] [B2])=(((5 + 2 x/2))2/((10 - x)2) ((15 - x)2)=((. 21 .)2/2 × 7)=31.5

Q. $10$ moles of $A_{2}$ , $15$ moles of $B_{2}$ and $5$ moles of $A B$ are placed in a $2$ litre vessel and allowed the come to equilibrium. The final concentration of $A B$ is $10.5 M$ ,

$A_{2} (g) + B_{2} (g) ⇌ 2 AB (g)$

Determine the value of equilibrium constant $(K_{C})$ for the reaction.

$25.3$

$31.5$

$36.3$

$40.5$

$\frac{5 + 2 x}{2} = 10.5$ $; x = 8$

$K_{C} = \frac{( [ A B ] ) ^{2}}{[ A _{2} ] [ B _{2} ]} = \frac{( \frac{5 + 2 x}{2} ) ^{2}}{( \frac{10 - x}{2} ) ( \frac{15 - x}{2} )} = \frac{( 21 ) ^{2}}{2 \times 7} = 31.5$

Q. 10 moles of A2​ , 15 moles of B2​ and 5 moles of AB are placed in a 2 litre vessel and allowed the come to equilibrium. The final concentration of AB is 10.5M , A2​( g)+B2​( g)⇌2AB(g) Determine the value of equilibrium constant (KC​) for the reaction.

25.3

31.5

36.3

40.5