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jarptica [38.1K]
3 years ago
5

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

s written under cellular conditions?
Chemistry
1 answer:
nikklg [1K]3 years ago
6 0

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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Answer:

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Explanation:

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3 years ago
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Answer : The correct rate law for the reaction is,

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For the given chemical equation:

CHCl_3(g)+Cl_2(g)\rightarrow CCl_2(g)+HCl(g)

Rate law expression for the reaction:

\text{Rate}=k[CHCl_3]^a[Cl_2]^b

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Expression for rate law for first observation:

0.0035=k(0.010)^a(0.010)^b ....(1)

Expression for rate law for second observation:

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Expression for rate law for third observation:

0.0098=k(0.020)^a(0.020)^b ....(3)

Expression for rate law for fourth observation:

0.027=k(0.040)^a(0.040)^b ....(4)

Dividing 1 from 2, we get:

\frac{0.0069}{0.0035}=\frac{k(0.020)^a(0.010)^b}{k(0.010)^a(0.010)^b}\\\\2=2^a\\a=1

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Calculation used :

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3 years ago
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