For the following reaction, Kp = 3.5 104 at 1495 K.
H2(g) + Br2(g) 2 HBr(g)
What is the value of Kp for the following reactions at 1495 K?
(a) HBr(g) 1/2 H2(g) + 1/2 Br2(g)
(b) 2 HBr(g) H2(g) + Br2(g)
(c) 1/2 H2(g) + 1/2 Br2(g) HBr(g)
a) K'p=(Kp)^{-1/2}=5.3*10^-3
b)K''p=(Kp)^{-1}=2.77*10^-5
c)K'''p=(Kp)^{1/2}=189.7
- Reversing the direction of a reaction gives the inverse of the equilibrium constant.
- When multiplying by a constant, raise K to the power of that constant
(a) HBr(g) ⇌ 1/2 H2(g) + 1/2 Br2(g)
Since this reaction is the reverse of the given reaction, we can use the fact that the value of Kp for a reverse reaction is the reciprocal of the original reaction.
Therefore, the value of Kp for reaction (a) at 1495 K is 1/3.5 * 10^4.
(b) 2 HBr(g) ⇌ H2(g) + Br2(g)
Since this reaction is just the given reaction written in reverse, the value of Kp for reaction (b) at 1495 K is the reciprocal of the original Kp value.
Therefore, the value of Kp for reaction (b) at 1495 K is 1/3.5 * 10^4.
(c) 1/2 H2(g) + 1/2 Br2(g) ⇌ HBr(g)
To find the Kp for reaction (c), we can use the fact that the value of Kp for a reaction is the reciprocal of its balanced equation.
Therefore, the value of Kp for reaction (c) at 1495 K is 1/(3.5 * 10^4).
To determine the value of Kp for each reaction, we can use the relationship between equilibrium constants and the coefficients of the balanced chemical equation.
(a) HBr(g) 1/2 H2(g) + 1/2 Br2(g)
Since the given reaction is the reverse of the original reaction, we can use the inverse of the original Kp value.
To find Kp for the reverse reaction, we take the reciprocal of the original Kp value:
Kp_reverse = 1/Kp = 1/(3.5 * 10^4)
Kp_reverse = 2.857 * 10^(-5)
Therefore, the value of Kp for the reaction HBr(g) 1/2 H2(g) + 1/2 Br2(g) at 1495 K is 2.857 * 10^(-5).
(b) 2 HBr(g) H2(g) + Br2(g)
To find Kp for this reaction, we use the coefficients of the balanced equation as exponents in Kp:
Kp = Kp_original^(coefficient_H2) * Kp_original^(coefficient_Br2)
Kp = Kp_original^(1) * Kp_original^(1)
Kp = Kp_original^2
Since we already know the value of Kp_original from the given information, we can calculate Kp for reaction (b):
Kp = (3.5 * 10^4)^2
Kp = 1.225 * 10^9
Therefore, the value of Kp for the reaction 2 HBr(g) H2(g) + Br2(g) at 1495 K is 1.225 * 10^9.
(c) 1/2 H2(g) + 1/2 Br2(g) HBr(g)
To determine Kp for this reaction, we can use the same approach as in reaction (b):
Kp = Kp_original^(coefficient_HBr)
Since the coefficient of HBr in the original equation is 1, we can directly use the value of Kp_original to find Kp for reaction (c):
Kp = (3.5 * 10^4)^1
Kp = 3.5 * 10^4
Therefore, the value of Kp for the reaction 1/2 H2(g) + 1/2 Br2(g) HBr(g) at 1495 K is 3.5 * 10^4.
For 1/2 the reaction, new k is sqrt Kp.
For 2 x reaction, new k is K^2p
For the reverse and 1/2 new k is 1/sqrt Kp.