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mr_godi [17]
4 years ago
15

Methanol burns in oxygen to form carbon dioxide and water. part a write a balanced equation for the combustion of methanol. writ

e a balanced equation for the combustion of methanol. ch3oh(l)+3o2(g)→co2(g)+3h2o(g) ch3oh(l)+o2(g)→co2(g)+2h2o(g) ch3oh(l)+2o2(g)→2co2(g)+4h2o(g) 2ch3oh(l)+3o2(g)→2co2(g)+4h2o(g) submitmy answersgive up part b calculate δh∘rxn at 25 ∘c. δh∘rxn = kj submitmy answersgive up part c calculate δs∘rxn at 25 ∘c. δs∘rxn = j/k submitmy answersgive up part d calculate δg∘rxn at 25 ∘c.
Chemistry
1 answer:
MakcuM [25]4 years ago
6 0
Part a)
Balanced equation:
2 CH₃OH(l) + 3 O₂(g) → 2 CO₂(g) + 4 H₂O(g)
Part b)
ΔH for the reaction at 25°C
ΔH = Enthalpy change of products - Enthalpy change of reactants
      = [(2 mol) * (-393.509 kJ/mol) + (4 mol)*(-241.83 kJ/mol)] - [(2 mol) * (238.4 kJ/mol) + (3 mol) *(0 kJ/mol)] = - 1277.5 kJ
Part c)
ΔS for the reaction at 25°C 
ΔS = entropy change in products - entropy change in reactants
     = [(2 x 213.7) + (4 x 188.8)] - [(2 x 127.2) + (3 x 205.1)] = 312.9 J K⁻¹
Part d)
ΔG for the reaction at 25°C
ΔG = ΔH - (T * ΔS)
     = - 1277.5 x 10³ J - (298 x 312.9 J K⁻¹) = -1370.7 kJ 
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Answer:

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\colorbox{red}Kp=Kc{[RT]}^{∆n}

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Given the molar absorptivity for a species X of 1600 M-1cm-1 at a wavelength of 270 nm, and 400 M-1cm-1 at a wavelength of 540 n
Andre45 [30]

Answer:

Explanation:

From the given information:

At wavelength = 270 nm

\varepsilon x_1 = 1600 \ m^{-1} \ cm^{-1}  \\ \\  \varepsilon y_1 = 200 \ m^{-1} \ cm^{-1}

At 270 nm

Suppose x is said to be the solution for the concentration of x and y to be the solution for the concentration of y;

Then:

\varepsilon x_1  \ l + \varepsilon y_1  \ l= 0.5 \\ \\ A = A_1 + A_2

1600 xl + 200 yl= 0.5

Divide both sides by 200

8xl + yl = \dfrac{0.5}{200}

8x + y = \dfrac{0.5}{200}l

Use l = 1cm (i.e the standard length)

Then;

8x + y = \dfrac{0.5}{200} ---- (1)

<u>For 540 nm:</u>

\varepsilon x_2 x  \ l + \varepsilon y_2 y  \ l= 0.5  \\ \\ 40 xl + 800 yl = 0.5

x + 20 y = \dfrac{0.5}{400 \ l}

since l = 1

x + 20 y = \dfrac{0.5}{400 \ } --- (2)

Equating both (1) and (2) together, we have:

8x + y - 8x - 160 y = \dfrac{0.5}{200} - \dfrac{0.5 \times 8}{400}  \\ \\  \implies - 159 y = \dfrac{0.5}{200} ( 1 - \dfrac{8}{2}) \\ \\  -159 y = \dfrac{-0.5 \times 3}{200}  \\ \\  159 \ y = 0.0075  \\ \\  y = \dfrac{0.0075}{159} \\ \\  y = 0.00004716 \\ \\ y = 4.7 \times 10^{-5 } \ M

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3 years ago
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