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puteri [66]
4 years ago
7

The equilibrium-constant expression depends on the of the reaction. (A) stoichiometry (B) mechanism (C) stoichiometry and mechan

ism (D) the quantities of reactions and products initially present (E) temperature
Chemistry
1 answer:
myrzilka [38]4 years ago
8 0

Answer:

(A) stoichiometry

Explanation:

For example, Consider the reaction of the type ,

aA + bB-----> cC +dD

where b, a, d, and c are the stoichiometric coefficients for the balanced equation .

The equilibrium constant expression is denoted by K ,

equilibrium constant is the product's concentration each raised to the power of its corresponding stoichiometries , divided by reactant's concentration each raised to the power of its corresponding stoichiometries .

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Given the following equation, N2O(g) + NO2(g) → 3 NO(g) ΔG°rxn = -23.0 kJ
Lelechka [254]

Answer:

ΔG°rxn = -69.0 kJ

Explanation:

Let's consider the following thermochemical equation.

N₂O(g) + NO₂(g) → 3 NO(g) ΔG°rxn = -23.0 kJ

Since ΔG°rxn < 0, this reaction is exergonic, that is, 23.0 kJ of energy are released. The Gibbs free energy is an extensive property, meaning that it depends on the amount of matter. Then, if we multiply the amount of matter by 3 (by multiplying the stoichiometric coefficients by 3), the ΔG°rxn will also be tripled.

3 N₂O(g) + 3 NO₂(g) → 9 NO(g) ΔG°rxn = -69.0 kJ

8 0
4 years ago
A particular reactant decomposes with a half‑life of 109 s when its initial concentration is 0.280 M. The same reactant decompos
Sophie [7]

Answer:

The order of reaction is 2.

Rate constant is 0.0328 (M s)⁻¹

Explanation:

The rate of a reaction is inversely proportional to the time taken for the reaction.

As we are decreasing the concentration of the reactant the half life is increasing.

a) For zero order reaction: the half life is directly proportional to initial concentration of reactant

b) for first order reaction: the half life is independent of the initial concentration.

c) higher order reaction: The relation between half life and rate of reaction is:

Rate = \frac{1}{k[A_{0}]^{(n-1)}}

Half life =K\frac{1}{[A_{0}]^{(n-1)} }

\frac{(halflife_{1})}{(halflife_{2})}=\frac{[A_{2}]^{(n-1)}}{[A_{1}]^{(n-1)} }

where n = order of reaction

Putting values

\frac{109}{231}=\frac{[0.132]^{(n-1)}}{[0.280]^{(n-1)}}

0.472=(0.472)^{(n-1)}

Hence n = 2

halflife=\frac{1}{k[A_{0}]}

Putting values

231=\frac{1}{K(0.132)}

K = 0.0328

4 0
3 years ago
PLEASE ANSWER ASAP AND SHOW WORK: A camping stove uses a 5.0 L propane tank that holds 68.0 moles of liquid C3H8. How large a co
seropon [69]

<u>Answer:</u> The volume of the container needed is 554.6 L

<u>Explanation:</u>

To calculate the volume of the gas, we use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 3.0 atm

V = Volume of the gas = ? L

T = Temperature of the gas = 25^oC=[25+273]K=298K

R = Gas constant = 0.0821\text{ L.atm }mol^{-1}K^{-1}

n = number of moles of propane gas = 68.0 moles

Putting values in above equation, we get:

3.0atm\times ?L=68mol\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\V=\frac{68\times 0.0821\times 298}{3.0}=554.6L

Hence, the volume of the container needed is 554.6 L

6 0
3 years ago
EZ POINTS!!!<br>what are 3 physical changes of wood.​
VARVARA [1.3K]
Answer: Cutting it, boiling it , freezing it
4 0
3 years ago
How many moles of O2 are needed to react<br> with 24 moles of C2H6?<br> 2C2H6 + 702 — 4CO2 + 6H20
Anna11 [10]

Explanation:

Mole ratio of Ethane to Oxygen = 2 : 7

Moles of O2 needed = 24 moles * (7/2) = 84 moles.

7 0
3 years ago
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