Answer:
Keq = 0.053
7.3 kJ/mol
Explanation:
Let's consider the following isomerization reaction.
glucose 6‑phosphate ⇄ glucose 1 - phosphate
The concentrations at equilibrium are:
[G6P] = 0.19 M
[G1P] = 0.01 M
The concentration equilibrium constant (Keq) is:
Keq = [G1P] / [G6P]
Keq = 0.01 / 0.19
Keq = 0.053
We can find the standard free energy change, ΔG°, of the reaction mixture using the following expression.
ΔG° = -R × T × lnKeq
ΔG° = -8.314 J/mol.K × 298 K × ln0.053
ΔG° = 7.3 × 10³ J/mol = 7.3 kJ/mol
The Law of Conservation of Mass states that matter cannot be created nor destroyed. The first two reactants have a total of 23 grams so therefore the product must have 23 grams as well.
23 - 16 = 7
There would be 7 grams of hydrogen gas.
Answer:
A
Explanation:
Combustion reaction always has something hydrocarbon plus oxygen gas which equals to carbon dioxide and water. Always being a reactant in a combustion reaction is oxygen, since reactants are what we start off the reaction with.
<u>Given:</u>
H2(g) + Cl2 (g) → 2HCl (g)
<u>To determine:</u>
The enthalpy of the reaction and whether it is endo or exothermic
<u>Explanation:</u>
Enthalpy of a reaction is given by the difference between the enthalpy of formation of reactants and products
ΔH = ∑nHf (products) - ∑nHf (reactants)
= [2Hf(HCl)] - [Hf(H2) + Hf(Cl2)] = 2 (-92.3) kJ = - 184.6 kJ
Since the reaction enthalpy is negative, the reaction is exothermic
<u>Ans:</u> The enthalpy of reaction is -184. kJ and the reaction is exothermic