Question

Homework/Chapter15 Problem 15.74 « previous 23 of 25 next » Problem 15.74 A sample of nitrosyl bromide (NOBr) decomposes according to the equation 2NOBr(g)⇌2NO(g)+Br2(g) An equilibrium mixture in a 5.00-L vessel at 100 ∘C contains 3.28 g of NOBr, 3.06 g of NO, and 8.15 g of Br2.

A.)Calculate Kc.

B.)What is the total pressure exerted by the mixture of gases?

C.)What was the mass of the original sample of NOBr?

Answer 1

Moles of NOBr present = 3.28/14+16+80 =.03 moles

Moles of NO present =3.06/14+16=.102 moles

Moles of Br2 present =8.17/2*80 =.051 moles

[NOBr] =.03/5=6*10^-3 M

[NO] =.102/5 =.0204 M

[Br2] =.051/5=.0102 M

1. Kc = [NO]²[Br2]/[NOBr]²

   =.0204²*.0102/(6*10^-3)²

          =.1133 M

2. Total no of moles of gas present =.03+.102+.051 =.183

Now, pV=nRT

=>p=.183*0.0820*(100+273)/5

    p= 1.12 atm

3. Since 2 moles of NOBr is consumed to produce 2 moles of NO

Moles of NOBr consumed to produce .102 moles of NO =.102

Therfore Initial moles of NOBr = .102 +.03 =.132

Therfore Initial mass of NOBr =.132*(14+16+80) = 14.52 g

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