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Dahasolnce [82]
4 years ago
11

Consider the following reactions. (Note: (s) = solid, (l) = liquid, and (g) = gas.)

Chemistry
2 answers:
Lelu [443]4 years ago
8 0

Answer:

Increase

Explanation:

A reaction can be exothermic or endothermic. An exothermic reaction, the heat flows from the system, means the system is losing heat, so the temperature of the system decreases and the temperature of the surroundings increases.

In an endothermic reaction, the heat flows to the system, so it's gain heat, the temperature of the system increase and the temperature of the surroundings decreases.

The enthalpy (H) is a measure of this heat, so for an exothermic reaction, it must decrease (because the system loses heat) and the variation ΔH must be negative. For an endothermic reaction, the opposite must occurs.

Then the reaction given is exothermic, the system loses heat and the temperature of the surroundings increase.

aalyn [17]4 years ago
6 0
Increase because the reaction is exothermic

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Which statement below correctly describes the relationship between Q and K for both reactions? Are these reactions spontaneous a
nikklg [1K]

Answer:

Q < K for both reactions. Both are spontaneous at those concentrations of substrate and product.

Explanation:

Hello,

In this case, the undergoing chemical reactions with their proper Gibbs free energy of reaction are:

A->B;\Delta _rG^o=-13 kJ/mol

C ->D ;\Delta _rG^o=3.5 kJ/mol

The cellular concentrations are as follows: [A] = 0.050 mM, [B] = 4.0 mM, [C] = 0.060 mM and [D] = 0.010 mM.

For each case, the reaction quotient is:

Q_1=\frac{4.0mM}{0.050mM}=80\\ Q_2=\frac{0.010mM}{0.060mM}=0.167

A typical temperature at a cell is about 30°C, in such a way, the equilibrium constants are:

K_1=exp(-\frac{-13000J/mol}{8.314J/mol*K*303.15K} )=173.8\\K_2=exp(-\frac{3500J/mol}{8.314J/mol*K*303.15K} )=0.249

Therefore, Q < K for both reactions. Both are spontaneous at those concentrations of substrate and product.

Best regards.

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