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tensa zangetsu [6.8K]
3 years ago
11

Acetylene, C2H2, can be converted to ethane, C2H6, by a process known as hydrogenation. The reaction is C2H2(g)+2H2(g)?C2H6(g)

Chemistry
2 answers:
saw5 [17]3 years ago
7 0

Answer:

part A =  K = 2.7*10^42

Part B =  ΔG  = -252284 J = -252.3 kJ

Explanation:

Step 1: Data given

Step 2: The balanced equation

C2H2(g)+2H2(g) ⇄ C2H6(g)

Δ Gf = (1* GfC2H6) - (1*GfC3H2)

Δ Gf = -209.2 -32.89

Δ Gf = - 242.09 kJ/mol  = -242090 J/mol

Step 3 Calculate K

Δ Gf = -RT ln K

⇒Δ Gf = -242090 J/mol

⇒R = 8.314 J/mol*K

⇒T = the temperature = 298 K

⇒ K = the equilibrium constant

-242090 = - (8.314)(298) lnK

K = 2.7*10^42

PArt B:

Δ Gf = -242.1 kJ= -242100 J

Q = pressure products / pressure reactants

Q = (pC2H6) / (pH2² * pC2H2)

Q = 1.25 / (4.45²*3.85)

Q= 0.0164

ΔG = ΔG° + RT ln Q

ΔG  = -242100  + 8.314 * 298 * ln 0.0164

ΔG  = -252284 J = -252.3 kJ

Alecsey [184]3 years ago
3 0

Answer:

Keq =1.50108

Explanation:

The given reactionis

 C₂H₂(g) +2H₂(g) -------------> C₂H₂(g)

   ΔG0 f=ΔG0f n (products) - ΔG0f n (reactants )

              = -32.89 kJ/mol - (209.2 kJ/mol+2*0.0 kJ/mol)

              = - 242.09kJ/mol

   ΔG= -RTlnKeq

   ln Keq = -ΔG/RT

           =-(- 242.09kJ/mol ) / 2 k cal /mol*298 K

           =0.406

    Keq =e0.406

         Keq =1.50108

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Answer:

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5 0
3 years ago
how much current would be measured in a circuit if the light bulb has a resistance of 6 ohms and a voltage of 36 volts
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Answer:

The right response is "6 A". A further explanation is given below.

Explanation:

The given values are:

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Voltage,

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As we know,

⇒  V=IR

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⇒  I=\frac{V}{R}

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Which has the highest boiling point .33 m NH3 or .10 m Na2SO4​
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Answer:

0.33 mol/kg NH₃

Explanation:

Data:

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b(Na₂SO₄) = 0.10 mol/ kg

Calculations:

The formula for the boiling point elevation ΔTb is

\Delta T_{b} = iK_{b}b

i is the van’t Hoff factor — the number of moles of particles you get from a solute.

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The ammonia is a weak electrolyte, so it exists almost entirely as molecules in solution.  

1 mol NH₃ ⟶  1 mol particles

i ≈ 1, and ib = 1 × 0.33 = 0.33 mol particles per kilogram of water

(b) For Na₂SO₄,

Na₂SO₄(aq) ⟶ 2Na⁺(aq) + 2SO₄²⁻(aq)

1 mol Na₂SO₄ ⟶ 3 mol particles

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3 years ago
Given the following equation,
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Answer:

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Explanation:

Given data:

Number of moles of HCl = 5 mol

Number of moles of H₂O produced = ?

Solution:

Chemical equation:

HCl + NaOH     →   NaCl + H₂O

Now we will compare the moles of HCl with H₂O.

                          HCl            :           H₂O

                             1               :             1

                             5               :            5

5 moles of water will be produced.

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