For reaction A(g) + B(g)leftharpoons AB(g) we start with 2 moles of A and B each. At equilibrium 0.8 moles of AB is formed. Then how much of A changes to AB

Question

For reaction A(g) + B(g)leftharpoons AB(g) we start with 2 moles of A and B each. At equilibrium 0.8 moles of AB is formed. Then how much of A changes to AB (A) 20 % (B) 40 % (C) 60 % (D) 4 %

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/f936864be9063496dae6838b3e808217-.png" /> ∴ % of A changed to A B =(0.8/2) × 100=40 %

Q. For reaction $A (g) + B (g) ⇌ A B (g)$ we start with $2$ moles of $A$ and $B$ each. At equilibrium $0.8$ moles of $A B$ is formed. Then how much of $A$ changes to $A B$

$20%$

$40%$

$60%$

$4%$

$∴ %$ of A changed to A B $= \frac{0.8}{2} \times 100 = 40%$

Q. For reaction A(g)+B(g)⇌AB(g) we start with 2 moles of A and B each. At equilibrium 0.8 moles of AB is formed. Then how much of A changes to AB

20%

40%

60%

4%