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gtnhenbr [62]
3 years ago
13

The average bond energy (enthalpy) for a C=C double bond is 614 kJ/mol and that of a C−C single bond is 348 kJ/mol. Estimate the

energy needed to break only the π bond of the double bond of 2-butene. Express your answer numerically in joules per molecule.
Chemistry
1 answer:
AysviL [449]3 years ago
7 0

Answer:

4.42x10⁻¹⁹ J/molecule

Explanation:

At a double bond, there's sigma and a pi bond, and at a single bond, there's only a sigma bond. Thus, if the energy to break both sigma and pi is 614 kJ/mol, and the energy to break only the sigma bond is 348 kJ/mol, the energy to break only the pi bond is:

E = 614 - 348 = 266 kJ/mol

Knowing that 1 kJ = 1000 J, E = 266,000 J/mol

By Avogadro's number, 1 mol = 6.02x10²³ molecules, thus:

E = 266,000 J/mol * 1mol/6.02x10²³ molecules

E = 4.42x10⁻¹⁹ J/molecule

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QveST [7]

<u>Answer:</u>

<u>For A:</u> The K_p for the given reaction is 4.0\times 10^1

<u>For B:</u> The K_c for the given reaction is 1642.

<u>Explanation:</u>

The given chemical reaction follows:

2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)

  • <u>For A:</u>

The expression of K_p for the above reaction follows:

K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}

We are given:

p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm

Putting values in above equation, we get:

K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1

Hence, the K_p for the given reaction is 4.0\times 10^1

  • <u>For B:</u>

Relation of K_p with K_c is given by the formula:

K_p=K_c(RT)^{\Delta ng}

where,

K_p = equilibrium constant in terms of partial pressure = 4.0\times 10^1

K_c = equilibrium constant in terms of concentration = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature = 500 K

\Delta ng = change in number of moles of gas particles = n_{products}-n_{reactants}=2-3=-1

Putting values in above equation, we get:

4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642

Hence, the K_c for the given reaction is 1642.

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neutral grain spirit as the name implies is neutral in nature that is it can be blended with any other alcoholic product.

In conclusion, The neutral grain spirit is essentially known as rectified spirits or ethyl alcohol.

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