At 450oC, the equilibrium constant Kc for the Haber-Bosch synthesis of ammonia is 0.16 for the reaction writeten as 3H2(g) + N2 (g)--><--2NH3(g) calculate the value of Kc for the same reaction written as 3/2H2(g) + 1/2N2(g)-->--<NH3(g)
3H2 + N2 ==> 2NH3 with Kc= 0.16
3/2H2 + 1/2N2 ==> NH3 is just 1/2 of the first equation; therefore, K'c = sqrt Kc or sqrt (0.16).
To calculate the value of Kc for the reaction written as 3/2H2(g) + 1/2N2(g) --> NH3(g), we can use the relationship between the stoichiometric coefficients and the equilibrium constant.
First, let's write the balanced equation for the reaction:
3/2H2(g) + 1/2N2(g) <=> NH3(g)
The stoichiometric coefficients in this equation are 3/2, 1/2, and 1, respectively.
Now, let's write the balanced equation for the original reaction:
3H2(g) + N2(g) <=> 2NH3(g)
The stoichiometric coefficients in this equation are 3, 1, and 2, respectively.
The relationship between the equilibrium constants for these two equations can be expressed as:
Kc2 = (Kc1)^(Δn)
Where Kc2 is the equilibrium constant for the second equation, Kc1 is the equilibrium constant for the first equation, and Δn is the change in the number of moles of gas molecules during the reaction.
In this case, Δn = (2 + 1) - (3/2 + 1/2) = 1 - 2 = -1
Plugging in the given value of Kc1 = 0.16 into the equation, we can calculate Kc2:
Kc2 = (0.16)^(-1)
To calculate the value, we take the reciprocal of Kc1 raised to the power of Δn.
Kc2 = 1 / 0.16
Kc2 = 6.25
Therefore, the value of Kc for the reaction written as 3/2H2(g) + 1/2N2(g) --> NH3(g) is 6.25.