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valkas [14]
3 years ago
11

For the reaction 2NH3(g) + 2O2(g)N2O(g) + 3H2O(l) H° = -683.1 kJ and S° = -365.6 J/K The standard free energy change for the rea

ction of 1.57 moles of NH3(g) at 273 K, 1 atm would be kJ. This reaction is (reactant, product) favored under standard conditions at 273 K. Assume that H° and S° are independent of temperature.
Chemistry
1 answer:
BlackZzzverrR [31]3 years ago
6 0

Answer:

\Delta G^{0} = -457.9 kJ and reaction is product favored.

Explanation:

The given reaction is associated with 2 moles of NH_{3}

Standard free energy change of the reaction (\Delta G^{0}) is given as:

           \Delta G^{0}=\Delta H^{0}-T\Delta S^{0}   , where T represents temperature in kelvin scale

So, \Delta G^{0}=(-683.1\times 10^{3})J-(273K\times -365.6J/K)=-583291.2J

So, for the reaction of 1.57 moles of NH_{3}, \Delta G^{0}=(\frac{1.57}{2})\times -583291.2J=-457883.592J=-457.9kJ

As, \Delta G^{0} is negative therefore reaction is product favored under standard condition.

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8 0
2 years ago
How many number of moles are present in 200. g of sodium bicarbonate (NaHCO3)
nikdorinn [45]

You can solve this problem through dimensional analysis.

First, find the molar mass of NaHCO3.

Na = 22.99 g

H = 1.008 g

C = 12.01 g

O (3) = 16 (3) g

Now, add them all together, you end with with the molar mass of NaHCO3.

22.99 + 1.008 + 12.01 + 16(3) = 84.008 g NaHCO3. This number means that for every mole of NaHCO3, there is 84.008 g NaHCO3. In simpler terms, 1 mole NaHCO3 = 84.008 g NaHCO3.

After finding the molar mass of sodium bicarbonate, now you can use dimensional analysis to solve for the number of moles present in 200. g of sodium bicarbonate.

200. g NaHCO_3 * \frac{1 mole NaHCO_3}{84.008 g NaHCO_3}

Cross out the repeating units which are g NaHCO3, and the remaining unit is mole NaHCO3

200.  * 1 = 200

200/ 84.008 = 2.38

Notice how there are only 3 sig figs in the answer. This is because the given problem only gave three sig figs.

Your final answer is 2.38 mol NaHCO3.

8 0
3 years ago
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Water molecules contain hydrogen atoms bonded directly to oxygen atoms. Oxygen atoms are highly electronegative and take major control of electrons in hydrogen-oxygen bonds. Hydrogen atoms in water molecules thus experience a strong partial-positive charge and would attract lone pairs of electron on neighboring water molecules. "Hydrogen bonds" refer to the attraction between hydrogen atoms bonded to electronegative elements and lone pairs of electrons. The hydrogen-oxygen bonds in water molecules are so polarized that hydrogen bonds in water are stronger than both dipole-dipole interactions and London Dispersion Forces in most other molecules. It thus take high amounts of energy to separate water molecules sufficiently apart such that they no longer experience intermolecular interactions and behave collectively like a gas. As a result, water has one of the highest heat of vaporization among covalent molecules of similar sizes.

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