Question

For the reaction
N2+3H2O---->2NH3 at 400K, KP=41
Find the value of Kp for 2N2+6H2----->4NH3

Answer 1

I think so in second reaction is it H2O.

If it is H2O answer is

1681.

Kp for reaction is =(NH3)*2÷(N2).(H2O)*3

Kp' for second reaction is =(NH3)*4÷(N2).(H2)*6

Kp' =( Kp')*2

Kp' = 41×41

= 1681.

Answer 2

Value of equilibrium constant for $2N_2+6H_2\rightarrow 4NH_3$ is 1681

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

The given chemical equation follows:

$N_2+3H_2\rightarrow 2NH_3$

The equilibrium constant for the above equation is 41.

We need to calculate the equilibrium constant for the reverse equation of above chemical equation, which is:

$2N_2+6H_2\rightarrow 4NH_3$

If the equation is multiplied by a factor of '2', the equilibrium constant of the reverse reaction will be the square of the equilibrium constant of initial reaction.

The value of equilibrium constant for reverse reaction is:

$K_{eq}'=(K_{eq})^2=(41)^2=1681$

Hence, the value of equilibrium constant for new reaction is 1681

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