Answer to Question #154917 in General Chemistry for pari

Question

At 585 K and a total pressure of 1.00 atm, NOCl(g) is 56.4/0 dissociated. Assume that 1.00  mol of ONCl(g) was present before dissociation, (a) How many moles of ONCl(g), NO(g),  and Cl2(g) are present at equilibrium? (b) What is the total number of moles of gas present  at equilibrium? (c) What are the equilibrium partial pressures of the three gases? (d) What  is the numerical value of Kp at 585 K? (The point is 4). 

Answer 1

2NOCl=2NO+Cl₂ 

(a) Let the initial amount of NOCl in the system be n moles. Then in the equilibrium state will remain:

$(1-a)\times n=0.436\times n$ mol NOCl

and at the same time, according to the chemical reaction equation, $a\times n=0.564\times n$ mol NO and $a\times \frac{n}{2}=0.282\times n$ mol Cl₂ are formed (according to stoichiometric coefficients).

The sum of the amounts of substances of the components in the system is $0.436\times n+0.564\times n+0.282\times n=1.282\times n$ moles.

n (NO)=0.564

n(Cl2)=0.282

n(NOCl)=0.436

(b) 1.282 moles

(c)1 atmosphere = 101325 Pa

Since the sum of the partial pressures of the components equals the total pressure in the system $p=pNOCl+pNO+pCl2=101325$ PA, and partial pressure of a component is proportional to the number of components in the system, find the partial pressure of component (n in number and the denominator of all formulas is reduced):

$pNOCl= 101325 \times\frac{0.436}{1.282}=34 459.98 Pa$

$pNO= 101325 \times\frac{0.564}{1.282}=44576.68 Pa$

$pCl2 = 101325\times\frac{0.282}{1.282}=22 288.34 Pa$

(d) The equilibrium constant for this reaction is written as:

$Kp=\frac{(pNOCl)^2}{((pNO)^2\times pCl2)}$ , respectively, equal to:

$Kp=\frac{34 459,98^2}{(44576.68^2\times22 288.34)}=0.000027$