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galben [10]
3 years ago
8

For the reaction: PCl5(g) ⇌ PCl3(g) + Cl2(g) at 600.0 K, the equilibrium constant is 11.5. Suppose that 1.500 g of PCl5 (MW=208.

22 g/mol) is placed in an evacuated 500.0 mL bulb, which is then heated to 600.0 K. What is the total pressure (in atm) in the bulb at equilibrium?
Chemistry
1 answer:
spayn [35]3 years ago
8 0

Answer:

1.418688 atm

Explanation:

(a)  Moles of PCl5 = mass / molar mass

                      =1.5 g / 208.22 g/mol

                     = 0.0072 moles

Also given,

T = 600 K

V = 0.500 L

Pressure of PCl5, P = nRT / V

                             = 0.0072 mol×0.0821 L-atm / (mol.K)×600 K / 0.500 L

                             = 0.709344 atm

(b)  PCl5(g) ⇄ PCl3(g) + Cl2(g)

Initial               0.965         0               0

Change               -x            +x             +x

Equilibrium (0.709344 -x)         x               x

K_p = 11.5 = x×x / (0.965 -x)

solving, we get x = 0.67027

So partial pressure of PCl5 at equilibrium = 0.709344 - 0.67027 = 0.039074 atm

(c)   Partial pressure of PCl3 = Cl2 = 0.709344 atm

So total pressure = 0.709344+0.039074+ 0.67027= 1.418688 atm

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The reaction between the reactants would be:

CH₃NH₂ + HCl ↔ CH₃NH₃⁺ + Cl⁻

Let the conjugate acid undergo hydrolysis. Then, apply the ICE approach.

             CH₃NH₃⁺ + H₂O → H₃O⁺ + CH₃NH₂
I                0.11                       0             0
C               -x                          +x           +x
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Since the given information is Kb, let's find Ka in terms of Kb.

Ka = Kw/Kb, where Kw = 10⁻¹⁴

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The expression for enthalpy change is,

\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]

Putting the values we get :

\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]

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2 moles of butane releases heat = 5314.8 kJ

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Thus enthalpy of combustion per mole of butane is -2657.4 kJ

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In case of diamond, the number of atoms is the same as the number of molecules.

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