What is the value for ∆soreaction for the following reaction, given the standard entropy values? 2h2s(g) + so2(g) 3srhombic(s) + 2h2o(g)

The answer is E for plato

Explanation:

#platolivesmatter

for the reaction is -186.75 J/K

Explanation:

Change in entropy () for the given reaction under standard condition is given by-

=

So = = -186.75 J/K

138 J/K

Explanation:

1) ΔS reaction = ΔS products - ΔS reactants

2) ΔS products = 2ΔS Fe(s) + 3 ΔS H₂O (g)

3) ΔS reactants = ΔS Fe₂O₃ (s) + 3 ΔS H₂ (g)

4) Now you have to search the standard entropy values for each product and each reactant in a table.

I found these values at 25° and 1 bar.

ΔS H₂O = 188.8 J/Kₓmol
ΔS Fe(s) = 27.3 J/Kmol

ΔS Fe₂O₃ (s) = 87.4 J/Kmol
ΔS H₂ (g) = 130.7 J/Kmol

5) Replace those values into above equations:

      ΔS products = 2ΔS Fe(s) + 3ΔS H₂O (g) = 
      2 mol × 27.3 J/K×mol + 3 mol ×188.8 J/K×mol = 621.0 J/K

      ΔS reactants = ΔS Fe₂O₃ (s) + 3 ΔS H₂ (g) =
      1mol × 87.4 J/K×mol + 3×130.7 J/K×mol = 479.5 J/K

      ΔS reaction = 621.0 J/K - 479.5 J/K = 141.5 J/K

Taking into account the differences in the values from different sources (specially due to the temperature), you can consider that the value 141.5 J/K is pretty much close to 138 J/K, and take that answer.

2Al(s) + 3Cl₂ → 2AlCl₃

Forming the equation:
2(28) + 3(223) = ΔH + 2(111)
ΔH = 503 kJ

Writing this outside of the reaction equation:
ΔH = -503 kJ

-503 J/K

Explanation:

The balanced chemical reaction is,

The expression for enthalpy change is,

where,

n = number of moles

Now put all the given values in this expression, we get

Therefore, the entropy change for this reaction is, -503J/K

141.55 J/K

Explanation: Value of ΔS can be obtained as:  

Standard entropy values are

Most likely a is the answer - because  you generate   38 atp's but also consume an atp.   each atp gives approx. less than 4 kcal so probably a is the closest answer.

The standard entropy for the substances are as follows:

C6H12O2(s) = -212 
O2(g) = -205 
CO2(g) = -214 
H2O(l) = -70

We calculate the ∆S°reaction by the expression:

∆S°rxn = ∆S°products - ∆S° reactants 
∆S°rxn = (212+6x205)-(6x214+6x70) 
∆S°rxn = -262 J/K > OPTION 3

To get the ∆S of the reaction, we simply have to add the ∆S of the reactants and the ∆S of the products. Then, we get the difference between the ∆S of the products and the ∆S of the products. If the ∆S is negative, then the reaction spontaneous. If the otherwise, the reaction is not spontaneous.

We know that the reaction equation is:
C(s) + O₂(g) → CO₂(g)

We may substitute the values of entropy and add a term of ΔS on the left hand side to represent the change in entropy. Thus:

6 + 205 + ΔS = 214
ΔS = 3 J/mol*K

The answer is C.