Chemistry, 29.07.2020 01:01 lakesha04begaye

Review the reversible reactions given, along with the associated equilibrium constant Kat room temperature. In each case, determine whether the forward or reverse reaction is favored. CH3COOH → CH3C00^- + H^+
Ka=1.8 x 10^-5
AgCl → Ag^+ + Cl^-
Ksp=1.6 x 10^-10
Al(OH)3 → Al^3+ + 3OH^-
Ksp=3.7 x 10^-15
A+B → C
K=4.9 x 10^3

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Answer from: wyattjefferds05

The answers to your questions are given below

Explanation:

The following data were obtained from the question:

CH3COOH → CH3C00^- + H^+

Equilibrium constant, Ka = 1.8 x 10^-5

AgCl → Ag^+ + Cl^-

Equilibrium constant, Ksp = 1.6 x 10^-10

Al(OH)3 → Al^3+ + 3OH^-

Equilibrium constant, Ksp = 3.7 x 10^-15

A+B → C

Equilibrium constant, K = 4.9 x 10^3

When the value of the equilibrium constant is grater than 1, it shows that the concentration of product is higher than that of the reactant and it implies that the forward reaction is favored.

When the value of the equilibrium constant is 1, it shows that the the concentration of the product and reactant are the same. Therefore neither the forward nor the reverse reaction is favored.

When the value of the equilibrium constant is lesser than 1, it shows that the concentration of the reactant is higher than the concentration of the product. Therefore, the reversed reaction is favored.

Now, we shall the question given above as follow:

A. CH3COOH → CH3C00^- + H^+

Equilibrium constant, Ka = 1.8 x 10^-5

Since the value of the equilibrium constant is lesser than 1, it means that the reverse reaction is favored.

B. AgCl → Ag^+ + Cl^-

Equilibrium constant, Ksp = 1.6 x 10^-10

Since the value of the equilibrium constant is lesser than 1, it means that the reverse reaction is favored.

C. Al(OH)3 → Al^3+ + 3OH^-

Equilibrium constant, Ksp = 3.7 x 10^-15

Since the value of the equilibrium constant is lesser than 1, it means that the reverse reaction is favored.

D. A+B → C

Equilibrium constant, K = 4.9 x 10^3

Since the value of the equilibrium constant is greater than 1, it means that the forward reaction is favored.

Answer from: Quest

11 players on a field that has a maximum of 11 players

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

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Review the reversible reactions given, along with the associated equilibrium constant Kat room tempe...