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Arte-miy333 [17]
3 years ago
7

sing any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 25.0°C for the following reaction

. C(s)+ 2Cl2(g)→ CCl4(g) Round your answer to 2 significant digits.
Chemistry
1 answer:
prisoha [69]3 years ago
3 0

Answer:

2.76 × 10⁻¹¹  

Explanation:

I don’t have access to the ALEKS Data resource, so I used a different source. The number may be different from yours.

1. Calculate the free energy of formation of CCl₄

                         C(s)+ 2Cl₂(g)→ CCl₄(g)

ΔG°/ mol·L⁻¹:       0         0         -65.3

ΔᵣG° = ΔG°f(products) - ΔG°f(reactants) = -65.3 kJ·mol⁻¹

2. Calculate K

\text{The relationship between $\Delta G^{\circ}$ and K  is}\\\Delta G^{\circ} = -RT \ln K

T = (25.0 + 273.15) K = 298.15 K

\begin{array}{rcl}-65 300 & = & -8.314 \times 298.15 \ln K \\65300& = & 2479 \ln K\\26.34 & = & \ln K\\K& = & e^{26.34}\\&= & \mathbf{2.76 \times 10}^{\mathbf{11}}\\\end{array}

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Please with this problem
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<h3>Answer:</h3>

            Empirical Formula  =  C₃H₈O₃

            Molecular Formula  =  C₃H₈O₃

<h3>Solution:</h3>

Data Given:

                      Mass of Sample  =  9.2 g

                      Mass of Carbon  =  3.6 g

                      Mass of Hydrogen  =  0.8 g

                      Mass of Oxygen  =  9.2 - (3.6 + 0.8) = 4.8 g

Step 1: Calculate Moles of each Element;

                      Moles of C  =  Mass of C ÷ At.Mass of C

                      Moles of C  = 3.6 ÷ 12.01

                      Moles of C  =  0.2997 mol


                      Moles of H  =  Mass of H ÷ At.Mass of H

                      Moles of H  = 0.8 ÷ 1.01

                      Moles of H  =  0.7920 mol


                      Moles of O  =  Mass of O ÷ At.Mass of O

                      Moles of O  = 4.8 ÷ 16.0

                      Moles of O  =  0.3000 mol

Step 2: Find out mole ratio and simplify it;

                C                                        H                                     O

            0.2997                              0.7920                           0.3000

    0.2997/0.2997                  0.7920/0.2997              0.3000/0.2997

               1                                      2.64                                    1.001

Multiply by 3,

               3                               7.92 ≈ 8                                    3

Hence,  Empirical Formula  =  C₃H₈O₃

Step 3: Calculating Molecular Formula:

Molecular formula is calculated by using following formula,

                    Molecular Formula  =  n × Empirical Formula  ---- (1)

Also, n is given as,

                     n  =  Molecular Weight / Empirical Formula Weight

Molecular Weight  =  92 g.mol⁻¹

Empirical Formula Weight  =  12 (C₃) + 1.01 (H₈) + 16 (O₃)  =  92.08 g.mol⁻¹

So,

                     n  =  92 g.mol⁻¹ ÷ 92 g.mol⁻¹

                     n  =  1

Putting Empirical Formula and value of "n" in equation 1,

                    Molecular Formula  = 1 × C₃H₈O₃

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8 0
4 years ago
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