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Anastasy [175]
3 years ago
12

When two gases of a chemical reaction are both at the same temperature and pressure and they have the same molar volume, what is

their stoichiometric ratio in the chemical equation describing the reaction?
Chemistry
2 answers:
Rudik [331]3 years ago
4 0
When two gases of a chemical reaction are at the same temperature, pressure and molar volume, then the stoichiometric ratio of the gases would be 1 is to 1. Molar volume is the volume of the gas per mole of the gas. Having the same conditions for both gases would mean that they are present with the same number of moles.
monitta3 years ago
3 0

Answer:

a. 1 to 1

Explanation:

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Consider the equilibrium
vladimir1956 [14]

Answer:

Kp^{1000K}=0.141\\Kp^{298.15K}=2.01x10^{-18}

\Delta _rG=1.01x10^5J/mol

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

C_2H_6(g)\rightleftharpoons H_2(g)+C_2H_4(g)

Thus, Kp for this reaction is computed based on the given molar fractions and the total pressure at equilibrium, as shown below:

p_{C_2H_6}^{EQ}=2bar*0.592=1.184bar\\p_{C_2H_4}^{EQ}=2bar*0.204=0.408bar\\p_{H_2}^{EQ}=2bar*0.204=0.408bar

Kp=\frac{p_{C_2H_4}^{EQ}p_{H_2}^{EQ}}{p_{C_2H_6}^{EQ}}=\frac{(0.408)(0.408)}{1.184}=0.141

Now, by using the Van't Hoff equation one computes the equilibrium constant at 298.15K assuming the enthalpy of reaction remains constant:

Ln(Kp^{298.15K})=Ln(Kp^{1000K})-\frac{\Delta _rH}{R}*(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=Ln(0.141)-\frac{137000J/mol}{8.314J/mol*K} *(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=-40.749\\\\Kp^{298.15K}=exp(-40.749)=2.01x10^{-18}

Finally, the Gibbs free energy for the reaction at 298.15K is:

\Delta _rG=-RTln(Kp^{298.15K})=8.314J/mol*K*298.15K*ln(2.01x10^{-18})\\\Delta _rG=1.01x10^5J/mol

Best regards.

3 0
4 years ago
What is the ph of a solution containing 0.01 m acetic acid (pka = 4.7) and 0.1 m sodium acetate?
Pepsi [2]

The ph of a solution is 3.7

Solution:

According to the equaiton of Henderson-Hasselback

      pH= pKa+ log(salt/acid)

here it is given the value of

      pKa= 4.7

So,

      pH = pKa+ log(0.1/0.01)

            = 4.7 + log(0.1)

            = 4.7–1

            = 3.7.

The following problem illustrates how the Henderson-Hasselbalch equation can be used to determine how much acid and conjugate base should be combined to create a buffer solution with a specific pH.

To learn more click the given link

brainly.com/question/13423434

#SPJ4

8 0
1 year ago
Which process in the light-dependent reactions results in the release of hydrogen ions, electrons, and oxygen?
tino4ka555 [31]
The answer is actually Water Splitting! Hope this works:D
6 0
3 years ago
Read 2 more answers
If 20. milliliters of 1.0 M HCl was used to completely neutralize 40. milliliters of an NaOH solution, what was the molarity of
olchik [2.2K]

Answer:

a)0.50

Explanation:

took one for the team :

5 0
3 years ago
DETERMINE THE MASS 1.366 mol of NH3
STALIN [3.7K]
Determine the mass of each of the following amounts. a. 1.366 mol of
NH3
, b. 0.120 mol of glucose,
C6H12O6
, c. 6.94 mol barium chloride,
BaCl2
, d. 0.005 mol of propane,
C3H8
4 0
3 years ago
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