The equilibrium constant for the following reaction is

Question

The equilibrium constant for the following reaction is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/c-rphp9lkfwb75xhcx.jpg" /> (A) 1011 (B) 10-11 (C) 1029 (D) 10-29

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/c-rphp9lkfwb75gswy.jpg" /> Subtracting (ii) from (i) gives the desired equation, Hence, equilibrium constant = (1 0- 9/1 0- 2 0) = 1 01 1 text.

Q. The equilibrium constant for the following reaction is

10¹¹

10^-11

10²⁹

10^-29

Subtracting (ii) from (i) gives the desired equation,
Hence, equilibrium constant $= \frac{1 0 ^{- 9}}{1 0 ^{- 20}} = 1 0^{11} .$

Q. The equilibrium constant for the following reaction is

1011

10-11

1029

10-29

Subtracting (ii) from (i) gives the desired equation, Hence, equilibrium constant =10−2010−9​=1011.

10¹¹

10^-11

10²⁹

10^-29

Subtracting (ii) from (i) gives the desired equation,
Hence, equilibrium constant $= \frac{1 0 ^{- 9}}{1 0 ^{- 20}} = 1 0^{11} .$