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guapka [62]
2 years ago
6

6. From the values of ΔH and ΔS, predict which of the following reactions would be spontaneous at 25ºC: Reaction A: ΔH = 10.5 kJ

/mol, ΔS = 30 J/K ∙ mol Reaction B: ΔH = 1.8 kJ/mol, ΔS = –113 J/K ∙ mol If any of the above reactions is nonspontaneous at 25ºC, at what temperature might it become spontaneous? (16 points)
Chemistry
1 answer:
qwelly [4]2 years ago
3 0

Answer:

Both reaction A and reaction B are non spontaneous.

Explanation:

For a spontaneous reaction, change in gibbs free energy (\Delta G) should be negative.

We know, \Delta G=\Delta H-T\Delta S, where T is temperature in Kelvin scale.

Reaction A: \Delta G=(10.5\times 10^{3})-(298\times 30)J/mol=1560J/mol

As \Delta G is positive therefore the reaction is non-spontaneous.

If at a temperature T K , the reaction is spontaneous then-

\Delta H-T\Delta S< 0

or, T> \frac{\Delta H}{\Delta S}

or, T> \frac{10.5\times 10^{3}}{30}

or, T> 350

So at a temperature greater than 350 K, the reaction is spontaneous.

Reaction B: \Delta G=(1.8\times 10^{3})-(-113\times 298)J/mol=35474J/mol

As \Delta G is positive therefore the reaction is non-spontaneous.

If at a temperature T K , the reaction is spontaneous then-

\Delta H-T\Delta S< 0

or, T> \frac{\Delta H}{\Delta S}

or, T> \frac{1.8\times 10^{3}}{-113}

or, T> -16

So at a temperature greater than -16 K, the reaction is spontaneous.

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Keith_Richards [23]

Answer:

\°C=5.8x10^3\°C

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Explanation:

Hello,

In this case, for the calculation of the temperature in degree Celsius we subtract 273.15 to the given temperature in kelvins:

\°C=6100-273.15\\\\\°C=5.8x10^3\°C

Next, by applying the following equation we compute it in degree Fahrenheit:

\°F=(5.8x10^{3}*9/5) + 32\\\\\°F=1.1x10^4\°F

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Regards.

5 0
2 years ago
The ph of a solution that contains 0.818 m acetic acid (ka = 1.76 x 10-5) and 0.172 m sodium acetate is __________.
crimeas [40]
PH is calculated using <span>Handerson- Hasselbalch equation,

                                      pH  =  pKa  +  log [conjugate base] / [acid]

Conjugate Base  =  Acetate (CH</span>₃COO⁻)
Acid                    =  Acetic acid (CH₃COOH)
So,
                                      pH  =  pKa  +  log [acetate] / [acetic acid]

We are having conc. of acid and acetate but missing with pKa,
pKa is calculated as,

                                     pKa  =  -log Ka
Putting value of Ka,
                                     pKa  =  -log 1.76 × 10⁻⁵

                                     pKa  =  4.75
Now,
Putting all values in eq. 1,
                     
                                     pH  =  4.75 + log [0.172] / [0.818]

                                     pH  =  4.072
4 0
2 years ago
Which balanced redox reaction is occurring in the voltaic cell represented by the notation of A l ( s ) | A l 3 ( a q ) | | P b
frez [133]

The question is missing. Here is the complete question.

Which balanced redox reaction is ocurring in the voltaic cell represented by the notation of Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}?

(a) Al_{(s)}+Pb^{2+}_{(aq)} ->Al^{3+}_{(aq)}+Pb_{(s)}

(b) 2Al^{3+}_{(aq)}+3Pb_{(s)} -> 2Al_{(s)}+3Pb^{2+}_{(aq)}

(c)Al^{3+}_{(aq)}+Pb_{(s)} ->Al_{(s)}+Pb^{2+}_{(aq)}

(d) 2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

Answer: (d) 2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

Explanation: <u>Redox</u> <u>Reaction</u> is an oxidation-reduction reaction that happens in the reagents. In this type of reaction, reagent changes its oxidation state: when it loses an electron, oxidation state increases, so it is oxidized; when receives an electron, oxidation state decreases, then it is reduced.

Redox reactions can be represented in shorthand form called <u>cell</u> <u>notation,</u> formed by: <em><u>left side</u></em> of the salt bridge (||), which is always the <em><u>anode</u></em>, i.e., its half-equation is as an <em><u>oxidation</u></em> and <em><u>right side</u></em>, which is always <em><u>the cathode</u></em>, i.e., its half-equation is always a <em><u>reduction</u></em>.

For the cell notation: Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)}

Aluminum's half-equation is oxidation:

Al_{(s)} -> Al^{3+}_{(aq)}+3e^{-}

For Lead, half-equation is reduction:

Pb^{2+}_{(aq)}+2e^{-} -> Pb_{(s)}

Multiply first half-equation for 2 and second half-equation by 3:

2Al_{(s)} -> 2Al^{3+}_{(aq)}+6e^{-}

3Pb^{2+}_{(aq)}+6e^{-} -> 3Pb_{(s)}

Adding them:

2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

The balanced redox reaction with cell notation Al_{(s)}|Al^{3+}_{(aq)}||Pb^{2+}_{(aq)}|Pb_{(s)} is

2Al_{(s)}+3Pb^{2+}_{(aq)} -> 2Al^{3+}_{(aq)}+3Pb_{(s)}

6 0
2 years ago
For each pair below, select the sample that contains the largest number of moles. Pair A 2.50 g O2 2.50 g N2
raketka [301]

Answer:

Explanation:

Pair  2.50g of O₂ and 2.50g of  N₂

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Number of moles of N₂ = \frac{2.5}{28} =  0.089mole

We see that N₂ has the largest number of moles

4 0
2 years ago
Read 2 more answers
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The answer is <span>D.when the aim is to show electron distributions in shells. This is because there are some instances when elements don't possess a regular or normal electron configuration. There are those who have special electron configurations wherein a lower subshell isn't completely filled before occupying a higher subshell. It is best to visualize such cases using the orbital notation.</span>
8 0
2 years ago
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