Question

The following reactions have the indicated equilibrium constants at a particular temperature: N2(g) + O2(g) ⇌ 2NO(g)Kc = 4.3 × 10−25 2NO(g) + O2(g) ⇌ 2NO2(g)Kc = 6.4 × 109 Determine the value of the equilibrium constant for the following equation at the same temperature: N2(g) + 2O2(g) ⇌ 2NO2(g) Kc = × 10 (Enter your answer in scientific notation.)

Answer 1

`K_c=2.752* 10^(-15)`

Step-by-step explanation:

The given equilibrium reaction is:

`N_2_((g))+O_2_((g))\rightleftharpoons 2NO_((g))`

The expression for the equilibrium constant is:

`K_c_1=\frac {[NO]^2}{[N_2][O_2]}=4.3* 10^(-25)`

The another given equilibrium reaction is:

`2NO{(g)}+O_2_((g))\rightleftharpoons 2NO_2_((g))`

The expression for the equilibrium constant is:

`K_c_2=\frac {[NO_2]^2}{[NO]^2[O_2]}=6.4* 10^(9)`

To find,

For the equilibrium which is:

`N_2_((g))+2O_2_((g))\rightleftharpoons 2NO_2_((g))`

The expression for the equilibrium constant is:

`K_c=\frac {[NO_2]^2}{[N_2][O_2]^2}`

Multiplying and dividing by
`[NO]^2` and rearranging in the above equation as:

`K_c=\frac {[NO_2]^2}{[NO]^2[O_2]}* \frac {[NO]^2}{[N_2][O_2]}`

`K_c=K_c_2* K_c_1=6.4* 10^(9)* 4.3* 10^(-25)=2.752* 10^(-15)`