I\'m needing help with #2 CHALLENGE PROBLEMS 1. Carbonyl bromide can dissociate

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I'm needing help with #2

CHALLENGE PROBLEMS 1. Carbonyl bromide can dissociate into carbon monoxide and bromine COBr2 (g) CO (9)+ Brz (9) If 17.91 g of pressureof carbonyl bromide is placed in a 10.37 L vessel and heated to 73 C, what is the partial pressure of carbon At 73°C, the equilibrium constant in terms of pressures, Kp, is 5.40 a. monoxide when equilibrium is attained? b. What fraction of carbonyl bromide is dissociated at equilibrium? 2. Consider the following equilibrium system. PCls (9) PCla (g)+ Cl2 (9) A 10.00 L evacuated flask is filled with 0.4493 mol PCls (g) at 297.9 K. Whe n the temperature is raised to 506.3 K, the decomposition of PCls occ 2.689 atm. What is the value of the equilibrium constant, Kc, at 506.3 K? 3. When excess Ag metal is added ot o 243 L of 1.320 M Fe (aq) at 298 K, the following equilibrium is urs. When equilibrium is established, the total pressure i established Ag (aq) + Fe* (aq) Ag (s) + Fe (aq) a. If the equilibrium concentration of Ag' is 0.517 M, what is the value of Ke? b. Using the value of Ke from part (a), calculate the new equilibrium concentrations of the three species in solution, if the volume of the solution is increased to 0.868 L by the addition of pure water?

Explanation / Answer

2)
Given:
V = 10.0 L
n = 0.4493 mol
T = 506.3 K

use:
P * V = n*R*T
P * 10 L = 0.4493 mol* 0.08206 atm.L/mol.K * 506.3 K
P = 1.867 atm

PCl5   <—>   PCl3    +   Cl2
1.867       0       0   (initial)
1.867-x   x       x   (at equilibrium)

total pressure = 1.867 - x + x + x = 1.867 +x
according to question,
1.867 +x = 2.689
x = 0.822

Kp = p(PCl3)p(Cl2)/p(PCl5)
= x*x/(1.867-x)
= (0.822*0.822)/(1.867-0.822)
= 0.647

T = 506.3 K
n = number of gaseous molecule in product - number of gaseous molecule in reactant
n = 1

use:
Kp= Kc (RT)^n
0.647 = Kc *(0.0821*506.3)^(1)
Kc = 1.56*10^-2
Answer: 1.56*10^-2

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