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StephyJ42
Mar 6, 2011, 06:47 PM
2. Consider the following reaction for which Kp = 0.0752 at 480.0 °C.
2 Cl2 (g) + 2 H2O (g) goes to 4 HCl (g) + O2 (g)
a) Calculate Kp for: 2 HCl (g) + ½ O2 (g) goes to Cl2 (g) + H2O (g)
b) Calculate Kp for: 8 Cl2 (g) + 8 H2O (g) goes to 16 HCl (g) + 4 O2 (g)




Kp = 0.0752 at 480.0 °C. 2Cl2(g) + 2H2O(g) goes to 4HCl(g) + O2(g)

Consider the following reaction for which Kp = 0.0752 at 480.0 °C.
2 Cl2 (g) + 2 H2O (g) goes to 4 HCl (g) + O2 (g)
Calculate Kc.

Unknown008
Mar 7, 2011, 06:53 AM
You have to know two things:

1. If a reaction of the type:

aA + bB \rightleftharpoons cC + dD

Then, the equation for Kp is:

K_{p1} = \frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}

And the reverse reaction is:

cC + dD \rightleftharpoons aA + bB

K_{p2} = \frac{(P_A)^a(P_B)^b}{(P_C)^c(P_D)^d}

In other words K_{p1} = \frac{1}{K_{p2}}

2. K constants are always always constant, except when temperature changes.

Also, K_c = \frac{[D]^d[C]^c}{[A]^a[B]^b}