The following reactions at 1000.0 K have the following K_p, values: H_2 (g) + Cl

Question

The following reactions at 1000.0 K have the following K_p, values: H_2 (g) + Cl_2 (g) 2 HCl (g) K_p = 3.8 10^H_2 (g) + Br_2 (g) + HBr (g) K_p = 5.1 10^Br_1 (g) + Cl_2 (g) 2 BrCl(g) K_p = 0.20 What it K_p, for the reaction. HCl(h) + HBr(g) BrCl(g) + H_2 (g) at 1000.0 K? Consider the following equilibrium: CO(g) + 3H_2 (g) CH (g) + H_2 O(g). If K_p = 1.61 10^at 1400.0 K_p calculate K.. a. 1.40 10^b. 1.85 10^c. 541 d. 0.212 e. 1.22 10^For the equilibrium 2 ClO(g) Cl_2 O_2 (g) K = 4.9610^at 253 K. What is K_p for the equilibrium 1/2 Cl_2 O_2 (g) ClO(g)? a. 7.04 10^b. 2.02 10^c. 9.90 10^d. 1.01 10^e. 1.42 10^HIO_3 behaves as acid in water. HIO_2 (aq) IO_2 (aq) + H'(aq), with K_p = 0.17 at 25C. What is the H' concentration in a solution that is initially 0.50 M HIO_4? a. 0.22 M b. 0.17 M c. 0.28 M d. 0.29 M e. 0.34 M

Explanation / Answer

Q5

Find HCl + HBr = BrCl + H2

we need HCl and HBr in the LEFT side so:

invert (1) and (2)

2HCl = H2 + Cl2 K = 1/(3.8*10^6)

2HBR = H2 + Br2 K = 1/(5.1*10^8)

let (3) as it is

Br2 + Cl2 = 2BrCl K = 0.20

add all

2HCl = H2 + Cl2 K = 1/(3.8*10^6)

2HBR = H2 + Br2   K = 1/(5.1*10^8)

Br2 + Cl2 = 2BrCl K = 0.20

2HCl +2HBr + Br2 + Cl2 =  H2 + Cl2+ H2 + Br2 +2BrCl K = 1/(3.8*10^6) * 1/(5.1*10^8) *0.20

cnacle common terms

2HCl +2HBr = 2HJ2 + 2BRCl K = 1.031*10^-16

but we need half so

HCl +HBr = H2 + 2BrCl K = (1.031*10^-16)^0.5 = 1.015*10^-8

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