1 mole of CH 3 COOH and 1 mole of C 2 H 5 OH reacts to produce (2/3) mole of CH 3 COOC 2 H 5. The equilibrium constant is

Question

1 mole of CH 3 COOH and 1 mole of C 2 H 5 OH reacts to produce (2/3) mole of CH 3 COOC 2 H 5. The equilibrium constant is (A) 2 (B) + 2 (C) -4 (D) + 4

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/0a4b4f986c87e35f8da2aad31b77760b-.png" /> Let the total volume = V L ∴ [ CH 3 OOH ]=(1/3) V mol / L [ C 2 H 5 OH ]=(1/3) V mol / L [ CH 3 COOC 2 H 5]=(2/3) V mol / L [ H 2 O ]=(2/3) V mol / L ∴ K =((2/3) V × (2/3) V /(1)3 V × (1)3 V =4

Q. 1 mole of CH3​COOH and 1 mole of C2​H5​OH reacts to produce 32​ mole of CH3​COOC2​H5​. The equilibrium constant is

2

+ 2

-4

+ 4

Let the total volume =VL ∴[CH3​OOH]=31​Vmol/L [C2​H5​OH]=31​Vmol/L [CH3​COOC2​H5​]=32​Vmol/L [H2​O]=32​Vmol/L ∴K=31​V×31​V32​V×32​V​=4

Q. $1$ mole of $C H_{3} COO H$ and $1$ mole of $C_{2} H_{5} O H$ reacts to produce $\frac{2}{3}$ mole of $C H_{3} COO C_{2} H_{5}$ . The equilibrium constant is