Hydroxyl radicals react with and eliminate many atmospheric pollutants. However, the hydroxyl radical does not clean up everythi
ng. For example, chlorofluorocarbons, which destroy stratospheric ozone, are not attacked by the hydroxyl radical. Consider the hypothetical reaction by which the hydroxyl radical might react with a chlorofluorocarbon:
OH(g)+CF2Cl2(g)→HOF(g)+CFCl2(g)
Use bond energies to explain why this reaction is improbable. Calculate ΔHrxn of this reaction.
OH(g)+CF2Cl2(g)→HOF(g)+CFCl2(g)
Use bond energies to explain why this reaction is improbable. Calculate ΔHrxn of this reaction.
1 answer:
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The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ of
) =-1275.0
enthalpy of combustion of oxygen(Δ of
) = zero
enthalpy of combustion of carbon dioxide(Δ of
) = -393.5
enthalpy of combustion of water(Δ of
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ
(products) - ∈Δ
(reactants)
(s) +6
(g) → 6
(g)+ 6
(l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Answer: The partial pressure of is 1.86 atm
Explanation:
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as
The given balanced equilibrium reaction is,
Pressure at eqm. 0.973 atm 0.548atm x atm
The expression for equilibrium constant for this reaction will be,
Now put all the given values in this expression, we get :
By solving the term 'x', we get :
x = 1.86 atm
Thus, the partial pressure of is 1.86 atm