Determine the value of the equilibrium constant for the reaction 2Br-(.aq.)+I2(.s.) leftharpoons Br2(.l.)+2I-(.aq.) Ec e l lo=-0.54 textV

Question

Determine the value of the equilibrium constant for the reaction 2Br-(.aq.)+I2(.s.) leftharpoons Br2(.l.)+2I-(.aq.) Ec e l lo=-0.54 textV (A) 5.01× 10- 19 (B) 18.30 (C) 1.7× 1054 (D) 1.9× 1018

Tardigrade Admin · Accepted Answer

The relation between equilibrium constant (K) and

E_{ce ll}^{o}

is given as

l o g K = \frac{E _{ce ll}^{o} . n}{0.059 V}

here, n = 2

= \frac{- 0.54 V \times 2}{0.059 V}

= \frac{- 1.08}{0.059} = - 18.3

On taking antilog

K = 5.01 \times 1 0^{- 19}

Q. Determine the value of the equilibrium constant for the reaction

$2 B r^{-} (a q) + I_{2} (s) ⇌ B r_{2} (l) + 2 I^{-} (a q)$

$E_{ce ll}^{o} = - 0.54 V$

$5.01 \times 1 0^{- 19}$

18.30

$1.7 \times 1 0^{54}$

$1.9 \times 1 0^{18}$

The relation between equilibrium constant (K) and $E_{ce ll}^{o}$ is given as

$l o g K = \frac{E _{ce ll}^{o} . n}{0.059 V}$

here, n = 2

$= \frac{- 0.54 V \times 2}{0.059 V}$

$= \frac{- 1.08}{0.059} = - 18.3$

On taking antilog

$K = 5.01 \times 1 0^{- 19}$

Q. Determine the value of the equilibrium constant for the reaction 2Br−(aq)+I2​(s)⇌Br2​(l)+2I−(aq) Ecello​=−0.54V

5.01×10−19

18.30

1.7×1054

1.9×1018

The relation between equilibrium constant (K) and Ecello​ is given as logK=0.059VEcello​.n​ here, n = 2 =0.059V−0.54V×2​ =0.059−1.08​=−18.3 On taking antilog K=5.01×10−19

Q. Determine the value of the equilibrium constant for the reaction

$2 B r^{-} (a q) + I_{2} (s) ⇌ B r_{2} (l) + 2 I^{-} (a q)$

$E_{ce ll}^{o} = - 0.54 V$

$5.01 \times 1 0^{- 19}$

$1.7 \times 1 0^{54}$

$1.9 \times 1 0^{18}$

The relation between equilibrium constant (K) and $E_{ce ll}^{o}$ is given as

$l o g K = \frac{E _{ce ll}^{o} . n}{0.059 V}$

here, n = 2

$= \frac{- 0.54 V \times 2}{0.059 V}$

$= \frac{- 1.08}{0.059} = - 18.3$

On taking antilog

$K = 5.01 \times 1 0^{- 19}$