Question

Consider the following equilibrium:

C₂H₆(g) + C₅H₁₂(g) <------> CH₄(g) + C₆H₁₄(g); K_p = 9.57 at 500 K

Suppose 54.2 g each of CH₄, C₂H₆, C₅H₁₂, and C₆H₁₄ are placed in a 46.0-L reaction vessel at 500 K. What is the value of Q_p?

A.9.57

B.1.00

C.0.104

D.1.57

E.0.637

Answer 1

1)

Molar mass of CH4 = 1*MM(C) + 4*MM(H)

= 1*12.01 + 4*1.008

= 16.042 g/mol

mass of CH4 = 54.2 g

we have below equation to be used:

number of mol of CH4,

n = mass of CH4/molar mass of CH4

=(54.2 g)/(16.042 g/mol)

= 3.379 mol

volume , V = 46.0 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 3.379/46

= 7.345*10^-2 M

2)

Molar mass of C6H14 = 6*MM(C) + 14*MM(H)

= 6*12.01 + 14*1.008

= 86.172 g/mol

mass of C6H14 = 54.2 g

we have below equation to be used:

number of mol of C6H14,

n = mass of C6H14/molar mass of C6H14

=(54.2 g)/(86.172 g/mol)

= 0.629 mol

volume , V = 46.0 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 0.629/46

= 1.367*10^-2 M

3)

Molar mass of C2H6 = 2*MM(C) + 6*MM(H)

= 2*12.01 + 6*1.008

= 30.068 g/mol

mass of C2H6 = 54.2 g

we have below equation to be used:

number of mol of C2H6,

n = mass of C2H6/molar mass of C2H6

=(54.2 g)/(30.068 g/mol)

= 1.803 mol

volume , V = 46.0 L

4)

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 1.803/46

= 3.919*10^-2 M

Molar mass of C5H12 = 5*MM(C) + 12*MM(H)

= 5*12.01 + 12*1.008

= 72.146 g/mol

mass of C5H12 = 54.2 g

we have below equation to be used:

number of mol of C5H12,

n = mass of C5H12/molar mass of C5H12

=(54.2 g)/(72.146 g/mol)

= 0.7513 mol

volume , V = 46.0 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 0.7513/46

= 1.633*10^-2 M

So, we have:

[CH4] = 7.345*10^-2 M

[C6H14] = 1.367*10^-2 M

[C2H6] = 3.919*10^-2 M

[C5H12] = 1.633*10^-2 M

Qc = [CH4][C6H14]/[C2H6][C5H12]

= (7.345*10^-2)*(1.367*10^-2)/((3.919*10^-2)*(1.633*10^-2))

= 1.57

Answer: 1.57