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Rama09 [41]
3 years ago
13

For the reaction c2h6 (g) → c2h4 (g) + h2 (g) δh° is +137 kj/mol and δs° is +120 j/k ∙ mol. this reaction is ________. question

10 options: spontaneous only at high temperature spontaneous at all temperatures spontaneous only at low temperature nonspontaneous at all temperatures
Chemistry
2 answers:
g100num [7]3 years ago
7 0
For the reaction c2h6 (g) → c2h4 (g) + h2 (g) δh° is +137 kj/mol and δs° is +120 j/k ∙ mol. This reaction is <span>spontaneous only at high temperature.
here the values of </span>δh° is +137 kj/mol and δs° is +120 kj/mol. so, both the value are wit the sign "+". when both values are positive than the reaction is not spontaneous at low temperature, the reaction is spontaneous only at high temperature.
tatyana61 [14]3 years ago
5 0

Answer: spontaneous only at high temperature

Explanation: C_2H_6(g)\rightarrow C_2H_4(g)+H_2(g)

The Gibbs free energy change is given by:

\Delta G=\Delta H-T\Delta S

\Delta G = Gibbs free energy

when \DeltaG= +ve, reaction is non spontaneous

\DeltaG= -ve, reaction is spontaneous

\DeltaG= 0, reaction is in equilibrium

\Delta H = enthalpy change = endothermic = +137 KJ

\Delta S = entropy change = +120 J/K

\Delta G=(+)-T(+)

\Delta G=(+)(-ve)

Now \Delta G= -ve when T\Delta S has more value than \Delta H

Thus reaction is spontaneous at high temperatures.

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In the combustion reaction CH4+2O2 yield CO2 + 2H2Owhich reactant has the greatest rate of disappearance?
Maksim231197 [3]

Answer:

To consume the 2.8 moles of CH4 we need 5.6 moles of O2 since the molar ratio is 1:2. We have only 3 moles of O2 ; therefore, O2 is the limiting reactant.

Explanation:

6 0
3 years ago
Thorium-234 has a half-life of 24.1 days. How many grams of a 150 g sample would you have after 120.5 days?
chubhunter [2.5K]

Answer:

8.625 grams of a 150 g sample of Thorium-234  would be left after 120.5 days

Explanation:

The nuclear half life represents the time taken for the initial amount of sample  to reduce into half of its mass.

We have given that the half life of thorium-234 is 24.1 days. Then it takes 24.1 days for a Thorium-234 sample to reduced to half of its initial amount.

Initial amount of Thorium-234 available as per the question is 150 grams

So now  we start with 150 grams  of Thorium-234

150 \times \frac{1}{2}=24.1

75 \times \frac{1}{2} =48.2

34.5 \times \frac{1}{2} =72.3

17.25 \times \frac{1}{2} =96.4

8.625\times \frac{1}{2} =120.5

So after 120.5 days the amount of sample that remains is 8.625g

In simpler way , we can use the below formula to find the sample left

A=A_{0} \cdot \frac{1}{2^{n}}

Where

A_0  is the initial sample amount  

n = the number of half-lives that pass in a given period of time.

7 0
3 years ago
What form of matter has a definite shape and takes up space?​
BARSIC [14]

Answer:

If something is in a solid state of matter, it has a definite shape and volume. The volume of an object is the amount of space it occupies. A block of wood placed on a table retains its shape and volume, therefore, it is an example of a solid. If a liquid is poured on that same table, there are very different results

Explanation:

7 0
2 years ago
Read 2 more answers
How many grams are in 32.2 L of CO2?
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Answer:

63.25 grams of CO₂

Explanation:

To convert from liters to grams, we first need to convert from liters to moles. To do this, we divide the liters by 22.4, the amount of liters of a gas per mole.

32.2 / 22.4

= 1.4375 moles of CO₂

Now we want to convert from moles to grams. To do this, we multiply the moles by the molar mass of CO₂. The total molar mass can be found on the periodic table by adding up the molar mass of carbon (12) and two oxygen (32).

12 + 32 = 44

Now we want to multiply the moles by the molar mass.

1.4375 • 44

= 63.25 grams of CO₂

This is your answer.

Hope this helps!

7 0
2 years ago
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Answer:

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