A 1 M solution of glucose reaches dissociation equilibrium according to equation given below 6HCHOleftharpoons C6H12O6. What is the concentration of HCHO at equilibrium if equilibrium constant is 6×1022

Question

A 1 M solution of glucose reaches dissociation equilibrium according to equation given below 6HCHOleftharpoons C6H12O6. What is the concentration of HCHO at equilibrium if equilibrium constant is 6×1022 (A) 1.6 × 10-8M (B) 32 × 10-6M (C) 3.2 × 10-4 M (D) 1.6 × 10-4M

Tardigrade Admin · Accepted Answer

6HCHO leftharpoons C6H12O6 forward reaction C6H12O6 leftharpoons 6HCHO backward reaction K2 =[(1/K1)]1/6 ;K2=[(1/6×1022)]1/6 K2 =1.6 × 10-4 M

Q. A $1 M$ solution of glucose reaches dissociation equilibrium according to equation given below $6 H C H O ⇌ C_{6} H_{12} O_{6}$ . What is the concentration of HCHO at equilibrium if equilibrium constant is $6 \times 1 0^{22}$

$1.6 \times 1 0^{- 8} M$

$32 \times 1 0^{- 6} M$

$3.2 \times 1 0^{- 4} M$

$1.6 \times 1 0^{- 4} M$

$6 H C H O ⇌ C_{6} H_{12} O_{6}$ forward reaction
$C_{6} H_{12} O_{6} ⇌ 6 H C H O$ backward reaction
$K_{2} = [\frac{1}{K _{1}}]^{1/6}; K_{2} = [\frac{1}{6 \times 1 0 ^{22}}]^{1/6}$
$K_{2} = 1.6 \times 1 0^{- 4} M$

Q. A 1M solution of glucose reaches dissociation equilibrium according to equation given below 6HCHO⇌C6​H12​O6​. What is the concentration of HCHO at equilibrium if equilibrium constant is 6×1022

1.6×10−8M

32×10−6M

3.2×10−4M

1.6×10−4M