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3 years ago

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For the following, identify the process as a spontaneous process, a nonspontaneous process, or an equilibrium process at the spe

cified temperature: (Assume that the thermodynamic data in the appendix do not vary with temperature.) CO2(g) → CO2(aq) at 25°C

1 answer:

alexandr402 [8]3 years ago

5 0

Answer:

A nonspontaneous process

Explanation:

The spontaneity of a reaction is given by the standard Gibbs free energy (ΔG°). We can calculate ΔG° using the following expression.

ΔG° = ∑np . ΔG°f(p) - ∑nr . ΔG°f(r)

where,

ni are the moles of reactants and products

ΔG°f(p) are the standard Gibbs free energies of formation of reactants and products

For the equation,

CO₂(g) → CO₂(aq)

ΔG° = 1 mol × ΔG°f(CO₂(aq)) - 1 mol × ΔG°f(CO₂(g))

ΔG° = 1 mol × (-386.0 kJ/mol) - 1 mol × (-394.4 kJ/mol)

ΔG° = 8.4 kJ

By convention, ΔG° > 0 means that the reaction is nonspontaneous.

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Sodium hydroxide, NaOH, reacts with sulfuric acid, H,SO,, in a neutralization reaction to

RSB [31]

Answer:

10.6 grams is approximately 0.10 moles. So you would need about 0.10 moles of sulfuric acid. That converts to about 9.80 grams.

Explanation:

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3 years ago

Hugh wrote the properties of physical and chemical weathering in the table shown.

Talja [164]

Answer:

I am sure that the C one is correct

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3 years ago

Read 2 more answers

A student dissolved 1.805g of a monoacidic weak base in 55mL of water. Calculate the equilibrium pH for the weak monoacidic base

yawa3891 [41]

Answer:

11.39

Explanation:

Given that:

$pK_{b}=4.82$

$K_{b}=10^{-4.82}=1.5136\times 10^{-5}$

Given that:

Mass = 1.805 g

Molar mass = 82.0343 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{1.805\ g}{82.0343\ g/mol}$

$Moles= 0.022\ moles$

Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity=\frac{0.022}{0.055}$

Concentration = 0.4 M

Consider the ICE take for the dissociation of the base as:

B + H₂O ⇄ BH⁺ + OH⁻

At t=0 0.4 - -

At t =equilibrium (0.4-x) x x

The expression for dissociation constant is:

$K_{b}=\frac {\left [ BH^{+} \right ]\left [ {OH}^- \right ]}{[B]}$

$1.5136\times 10^{-5}=\frac {x^2}{0.4-x}$

x is very small, so (0.4 - x) ≅ 0.4

Solving for x, we get:

x = 2.4606×10⁻³ M

pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61

<u>pH = 14 - pOH = 14 - 2.61 = 11.39</u>

5 0

3 years ago

Help..................................................................:)

34kurt

Answer:

ppppp

Explanation:

4 0

3 years ago

At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react

erastova [34]

Answer:

At equilibrium, the concentration of $N_{2 (g)}$ is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

$N_{2 (g)}$ + $O_{2 (g)}$ ⇄ 2 $NO_{(g)}$

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no $N_{2 (g)}$ nor $O_{2 (g)}$ present. Immediately, $N_{2 (g)}$ and $O_{2 (g)}$ are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [ $N_{2 (g)}$ ]=0 ; [ $O_{2 (g)}$ ]= 0 ; [ $NO_{(g)}$ ]=0.60M

C: [ $N_{2 (g)}$ ]=+x ; [ $O_{2 (g)}$ ]= +x ; [ $NO_{(g)}$ ]=-2x

E: [ $N_{2 (g)}$ ]=0+x ; [ $O_{2 (g)}$ ]= 0+x ; [ $NO_{(g)}$ ]=0.60-2x

Now we can use the constant information:

$K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }$

$4.10* 10^{-4} =\frac{(0.60-2x)^{2}}{(x)*(x)}$

$4.10* 10^{-4}$ = $\frac{(0.60-2x)^{2}}{x^{2} }$

$4.10* 10^{-4} * x^{2}$ = $(0.60-2x)^{2}}$

$\sqrt{4.10* 10^{-4} * x^{2}}$ = $\sqrt{(0.60-2x)^{2}}}$

$0.0202 x =0.60 - 2x$

$2x+0.0202x=0.60$

$x=\frac{0.60}{2.0202}$ $= 0.30$

At equilibrium, the concentration of $N_{2 (g)}$ is going to be 0.30M

3 0

3 years ago