Question

Consider the following generic reaction for which Kp = 5.51 × 105 at 25°C:

Answer 1

Answer:

$K_c=1.35\times 10^7$

Explanation:

The relation between Kp and Kc is given below:

$K_p= K_c\times (RT)^{\Delta n}$

Where,  

Kp is the pressure equilibrium constant

Kc is the molar equilibrium constant

R is gas constant

T is the temperature in Kelvins

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)

For the first equilibrium reaction:

$2R_{(g)}+A_{(g)}\rightleftharpoons2Z_{(g)}$

Given: Kp = $5.51\times 10^5$  

Temperature =  25°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

T = (25 + 273.15) K = 298.15 K  

R = 0.082057 L atm.mol⁻¹K⁻¹

Δn = (2)-(2+1) = -1  

Thus, Kc is:

$5.51\times 10^5= K_c\times (0.082057\times 299)^{-1}$

$K_c\frac{1}{24.535043}=551000$

$K_c=13518808.69=1.35\times 10^7$