Chemistry, 21.11.2019 04:31 juju323261

Acritical reaction in the production of energy to do work or drive chemical reactions in biological systems is the hydrolysis of adenosine triphosphate, atp, to adenosine diphosphate, adp, as described by the reaction atp(aq)+h2o(l)⟶adp(aq)+hpo2−4(aq) for which δ∘rxn=−30.5 kj/mol at 37.0 °c and ph 7.0. calculate the value of δ in a biological cell in which [atp]=5.0 mm, [adp]=0.70 mm, and [hpo2−4]=5.0 mm.

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Answer from: kyra737

Answer : The value of $\Delta G_{rxn}$ is -49.2 kJ/mol

Explanation :

First we have to calculate the reaction quotient.

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given balanced chemical reaction is,

$ATP(aq)+H_2O(l)\rightarrow ADP(aq)+HPO_4^{2-}(aq)$

The expression for reaction quotient will be :

$Q=\frac{[ADP][HPO_4^{2-}]}{[ATP]}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Given:

[ATP] = 5.0 mM

[ADP] = 0.70 mM

$[HPO_4^{2-}]$ = 5.0 mM

Now put all the given values in this expression, we get

$Q=\frac{(0.70)\times (5.0)}{(5.0)}=0.70mM=0.70\times 10^{-3}M$

Now we have to calculate the value of $\Delta G_{rxn}$ .

The formula used for $\Delta G_{rxn}$ is:

$\Delta G_{rxn}=\Delta G^o+RT\ln Q$ ............(1)

where,

$\Delta G_{rxn}$ = Gibbs free energy for the reaction = ?

$\Delta G_^o$ = standard Gibbs free energy = -30.5 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = 37.0^oC=273+37.0=310K

Q = reaction quotient = $0.70\times 10^{-3}$

Now put all the given values in the above formula 1, we get:

$\Delta G_{rxn}=(-30.5kJ/mol)+[(8.314\times 10^{-3}kJ/mole.K)\times (310K)\times \ln (0.70\times 10^{-3})$

$\Delta G_{rxn}=-49.2kJ/mol$

Therefore, the value of $\Delta G_{rxn}$ is -49.2 kJ/mol

Answer from: Quest

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form( $n_2$ ) .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

Answer from: Quest

possibly, yes.

explanation:

the amount of sugar you take in is dangerously high, risk of heart disease, high risk of diabetes, and etc.

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Acritical reaction in the production of energy to do work or drive chemical reactions in biological...