Problem 9.123 When 1.90 mol of NOCl( g ) was heated to 225 C in a 386.0 L steel

Question

Problem 9.123

When 1.90 mol of NOCl(g) was heated to 225 C in a 386.0 L steel reaction vessel, the NOClpartially decomposed according to the equation:
2NOCl(g)2NO(g)+Cl2(g).
The pressure in the vessel after reaction is 0.240 atm .

Part A

What is the partial pressure of each gas in the vessel after reaction?

Express your answers numerically separated by commas.

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Part B

What percent of the NOCldecomposed?

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Problem 9.123

When 1.90 mol of NOCl(g) was heated to 225 C in a 386.0 L steel reaction vessel, the NOClpartially decomposed according to the equation:
2NOCl(g)2NO(g)+Cl2(g).
The pressure in the vessel after reaction is 0.240 atm .

Part A

What is the partial pressure of each gas in the vessel after reaction?

Express your answers numerically separated by commas.

PNO, PCl2, PNOCl2= atm

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Part B

What percent of the NOCldecomposed?

percent NOCl= %

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Explanation / Answer

A)

What is the partial pressure of each gas in the vessel after reaction?

When 1.90 mol of NOCl(g) was heated to 225 C in a 386.0 L steel reaction vessel, the NOClpartially decomposed according to the equation: 2NOCl(g)2NO(g)+Cl2(g). The pressure in the vessel after reaction is 0.240 atm .

initial P

PV = nRT

P = nRT/V = (1.9*0.082*(225+273))/386 = 0.201 atm

initially

PNOCl = 0.201

PNO = 0

PCl2 = 0

after reaction

PNOCl = 0.201 - 2x

PNO = 0+2x

PCl2 = 0 + x

Pfinal = 0.240

0.201 - 2x + 0+2x +  0 + x = 0.240

x = 0.24 -0.201 = 0.039

PNOCl = 0.201 - 2x = 0.201 -2*0.039 = 0.123 atm

PNO = 0+2x = 2*0.039 = 0.078 atm

PCl2 = 0 + x = 0.039 atm

B)

% decomposition = (0.24-0.123) /(0.24) *100% = 48.75% has decomposed

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