Question

At 5000C the reaction N2+3H2 =2NH3.has Kc= 6.0×10 -2 . Calculate the value of Kp, where R=0.0821.​

Answer 1

Answer:

The value of  $K_{p}$  for the given reaction is $1.49$ × $10^{-5}$·

Explanation:

For the reaction,

         $N_{2} + 3 H_{2} = 2NH_{3}$

Given,

     $K_{c} = 6.0$ × $10^{-2}$

     T   $= 500^{0C} = 500+273 = 773K$

     R   $= 0.0821L atm/mol K$

To find $K_{p}$,

The relationship between $K_{p}$ and $K_{c}$ is,

        $K_{p} = K_{c}( RT)^{\Delta n}$

where,

   Δn $=$  No. of. moles of gaseous product $-$ No. of. moles of gaseous               reactants

⇒  Δn  $= 2-(3+1) = 2-4 = -2$

 On substituting the values,

   we get,

 ⇒    $K_{p} = 6.0$ × $10^{-2} ( 0.0821$ × $773)^{-2}$

 ⇒    $K_{p} = 6.0$ × $10^{-2}$ × $2.48$ × $10^{-4}$

 ⇒    $K_{p}= 1.49$ × $10^{-5}$