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Zigmanuir [339]
2 years ago
13

In the first step of glycolysis, the given two reactions are coupled. reaction 1:reaction 2:glucose+Pi⟶glucose-6-phosphate+H2OAT

P+H2O⟶ADP+PiΔG=+13.8 kJ/molΔG=−30.5 kJ/mol Answer the four questions about the first step of glycolysis. Is reaction 1 spontaneous or nonspontaneous?
Chemistry
1 answer:
dmitriy555 [2]2 years ago
6 0

Answer: Reaction 1 is non spontaneous.

Explanation:

According to Gibb's equation:

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

\Delta G = Gibbs free energy  

\Delta H = enthalpy change

\Delta S = entropy change  

T = temperature in Kelvin

When \Delta G = +ve, reaction is non spontaneous

\Delta G= -ve, reaction is spontaneous

\Delta G= 0, reaction is in equilibrium

For the given reaction 1: glucose+Pi\rightarrow glucose-6-phosphate+H_2O \Delta G=+13.8kJ/mol

As for the reaction 1 , the value of Gibbs free energy is positive and thus the reaction 1 is non spontaneous.

You might be interested in
Octane (C8H18) undergoes combustion according to the following thermochemical equation. 2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(l
Zepler [3.9K]

Answer: The standard enthalpy of formation of liquid octane is -250.2 kJ/mol

Explanation:

The given balanced chemical reaction is,

2C_8H_{18}(l)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(l)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{O_2}\times \Delta H_f^0_{(O_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})+n_{O_2}\times \Delta H_f^0_{(O_2)}]

where,

We are given:

\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(C_8H_{18}(l))}=?kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol

Putting values in above equation, we get:

-1.0940\times 10^4=[(16\times -393.5)+(18\times -285.8)]-[(25\times 0)+(2\times \Delat H_f{C_8H_{18}(l)}]

\Delta H^o_f_{(C_8H_{18}(l))}=-250.2kJ/mol

Thus the standard enthalpy of formation of liquid octane is -250.2 kJ/mol

4 0
2 years ago
Select all true statements about hybridization, of a second-row element, changing from sp2 to sp3. (max of 3 choices) a. The ato
Neko [114]

Answer:

a. The atom will go from a two-dimensional configuration to a three dimensional configuration.

d. The bond angle will increase.

f. The number of unhybridized p orbitals will decrease.

Explanation:

Sp2 is the atomic bond in which orbitals mixes with only two orbitals. These orbitals form three sp2. When two carbon atoms are overlapped they form sigma bond by overlapping of sp2 bonds. Sp3 bond is created when there is one lone molecule available for combination. When the bonding is updated from sp2 to sp3 then unhybridized orbitals will decrease causing the bond angle to increase.

8 0
2 years ago
Which statement explains why NaBr is classified as a compound?
charle [14.2K]

Answer:1

Explanation:i know cuz I got it right

5 0
2 years ago
trans-2-Butene does not exhibit a signal in the double-bond region of the spectrum (1600–1850 cm-1); however, IR spectroscopy is
spayn [35]

Answer:

The other signal that would indicate the presence of a C= C bond appears close to 3100 cm^{-1}.

Explanation:

Bands that appear above 3000 cm^{-1}  are often unsaturation diagnoses suggest. The band at 3000- 3100 cm^{-1} is characteristics for C-H stretching frequencies and normally is overlaps with the ones for alkanes because it is a band of weak intensity.

4 0
2 years ago
2. The diagram below shows steps in the exothermic chemical reaction of bromomethane with
vovangra [49]

Answer:

Here's what I get  

Explanation:

1. Balanced equation

HQ⁻ + CH₃-Br ⟶ HQ-CH₃ + Br⁻

(I must use HQ because the Brainly Editor thinks the O makes a forbidden word)

2. Mechanism

HQ⁻ + CH₃-Br ⟶[HQ···CH₃···Br]⁻⟶ HQ-CH₃ + Br⁻

        A                           C                         B

The hydroxide ion attacks the back side of the carbon atom in the bromomethane (A).

At the same time as the Q-H bond starts to form, the C-Br bond starts to break.

At the half-way point, we have a high-energy intermediate (C) with partially formed C-O and C-Br bonds.

As the reaction proceeds further, the Br atom drops off to form the products — methanol and bromide ion (B).

3. Energy diagram

See the diagram below.

 

8 0
2 years ago
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