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3 years ago

What part of a potential energy diagram reveals the enthalpy of reaction?

Chemistry

2 answers:

riadik2000 [5.3K]3 years ago

8 0

The difference in potential energy between the reactants and

products is the part of a potential energy diagram that reveals the enthalpy of reaction.

saveliy_v [14]3 years ago

7 0

Answer:

The difference in potential energy between the reactants and products

Explanation:

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Given the following information, what is the concentration of H2O(g) at equilibrium? [H2S](eq) = 0.671 M [O2](eq) = 0.587 M Kc =

MAVERICK [17]

Answer: The equilibrium concentration of water is 0.597 M

Explanation:

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{c}$

For a general chemical reaction:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_{eq}$ is written as:

$K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

The concentration of pure solids and pure liquids are taken as 1 in the expression.

For the given chemical reaction:

$2H_2S(g)+O_2(g)\rightleftharpoons 2S(s)+2H_2O(g)$

The expression of $K_c$ for above equation is:

$K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}$

We are given:

$[H_2S]_{eq}=0.671M$

$[O_2]_{eq}=0.587M$

$K_c=1.35$

Putting values in above expression, we get:

$1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}$

$[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M$

Hence, the equilibrium concentration of water is 0.597 M

8 0

4 years ago

(6.022*10^23)*2.58= what’s the answer

Sladkaya [172]

The answer would be
$1.553676 \times {10}^{24}$

8 0

4 years ago

Read 2 more answers

A concentration cell consisting of two hydrogen electrodes (PH2 = 1 atm), where the cathode is a standard hydrogen electrode and

Lunna [17]

The pH of the unknown solution is 3.07.

Explanation:

1.Find the cell potential as a function of pH

From the Nernst Equation:

Ecell=E∘cell−RT /zF × lnQ

where

R denotes the Universal Gas Constant

T denotes the temperature

z denotes the moles of electrons transferred per mole of hydrogen

F denotes the Faraday constant

Q denotes the reaction quotient

Substitute the values,

E∘cell=0 lnQ=2.303logQ

E0cell=−2.30/RT /zF × log Q

Solving the equation,

2. Find the Q value

Q=[H+]2prod pH₂, product/ [H+]2reactpH₂, reactant

Q=[H+]^2×1/1×1=[H+]2

Taking the log

logQ= log[H+]^2=2log[H+]=-2pH

From the formula,

Ecell=−2.303RT /zF× logQ

E cell= 2.303 × 8.314 CK mol (inverse) × 298.15

K × 2pH /2×96 485 C⋅mol

( inverse)

E cell= 0.0592 V × pH

3. Finding the pH value

E cell= 0.0592 V × pH

pH = E cell/ 0.0592 V= 0.182V/ 0.0592V

pH=3.07

The pH of the unknown solution is 3.07.

7 0

3 years ago

suffixes Refer to the word parts table that you made for the words in Section 2 containing the suffix -ic. Use each of the words

Sladkaya [172]

Explanation:

need an image

3 0

3 years ago

A solution contains the ions ag+ , pb2+ and ni2+. dilute solutions of nacl, na2so4 and na2s are available to separate the positi

sladkih [1.3K]

Use the sequence E (NaCl, Na2SO4, then Na2S). Silver is insoluble as a chloride, so it would be removed first, the others (Pb and Ni) are soluble as chlorides(Note; lead chloride is soluble as a hot solution but will ppt when cold), next, PbSO4 is insoluble but NiSO4 is soluble so use Na2SO4 to separate lead from nickel. Lastly, nickel sulfide is insoluble and can be separated and collected.

Hope I helped :)

6 0

3 years ago

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