Question

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

2R(g)+A(g)⇌2Z(g)

What is the value of Kc for the reaction at the same temperature?

1.37 × 10^9
2.25 × 10^4
1.35 × 10^7
5.51 × 10^5

Answer 1

`K_c=1.35* 10^7`

Step-by-step explanation:

The relation between Kp and Kc is given below:

`K_p= K_c* (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* 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* 10^5= K_c* (0.082057* 299)^(-1)`

`K_c(1)/(24.535043)=551000`

`K_c=13518808.69=1.35* 10^7`