Question

use the references to access important values if needed for th
the equilibrium constant, ( k_{p} ), for the following reaction is 0.0180 at 698 k.
calculate ( k_{c} ) for this reaction at this temperature.
2 mathrm{hi}(g)
ightarrow mathrm{h}_{2}(g)+mathrm{i}_{2}(g)
( k_{c}= )
submit answer retry entire group 6 more group attempts remaining

Explanation:

Step1: Determine the formula relating \(K_p\) and \(K_c\)

The formula is \(K_p = K_c(RT)^{\Delta n}\), where \(R = 0.0821\ L\cdot atm/(mol\cdot K)\), \(T\) is the temperature in Kelvin, and \(\Delta n\) is the change in the number of moles of gas.
For the reaction \(2HI(g)\to H_2(g)+I_2(g)\), \(\Delta n=(1 + 1)-2=0\).

Step2: Substitute the values into the formula

Since \(\Delta n = 0\), then \((RT)^{\Delta n}=(RT)^0 = 1\).
Given \(K_p = 0.0180\), and using \(K_p = K_c(RT)^{\Delta n}\), we have \(K_c=\frac{K_p}{(RT)^{\Delta n}}\).
Substituting \(\Delta n = 0\) and \(K_p=0.0180\) into the formula, we get \(K_c = K_p\) (because \((RT)^0 = 1\)).

Answer:

\(0.0180\)