1) Write the balanced equation to state the molar ratios:
<span>3H2(g) + N2(g) → 2NH3(g)
=> molar ratios = 3 mol H2 : 1 mol N2 : 2 mol NH3
What volume of nitrogen is needed to produce 250.0 L of ammonia gas at STP?
First, convert the 250.0 L of NH3 to number of moles at STP .
Use the fact that 1 mole of gas at STP occupies 22.4 L
=> 250.0 L * 1mol/22.4 L = 11.16 L
Second, use the molar ratio to find the number of moles of N2 that produces 11.16 L of NH3
=> 11.16 L NH3 * [1 mol N2 / 2 mol NH3] = 5.58 mol N2
Third, convert 5.58 mol N2 into liters at STP
=> 5.58 mol N2 * [22.4 L/mol] = 124.99 liters
Answer: 124,99 liters
What volume of hydrogen is needed to produce 2.50 mol NH3 at STP?
First, find the number of moles of H2 that produce 2.50 mol by using the molar ratios:
2.50 mol NH3 * [3mol H2 / 2 mol NH3] = 3.75 mol H2
Second, convert the number of moles to liters of gas at STP:
3.75 mol * 22.4 L/mol = 84 liters of H2
Answer: 84 liters
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Answer:
Kc = 8.05x10⁻³
Explanation:
This is the equilibrium:
2NH₃(g) ⇄ N₂(g) + 3H₂(g)
Initially 0.0733
React 0.0733α α/2 3/2α
Eq 0.0733 - 0.0733α α/2 0.103
We introduced 0.0733 moles of ammonia, initially. So in the reaction "α" amount react, as the ratio is 2:1, and 2:3, we can know the moles that formed products.
Now we were told that in equilibrum we have a [H₂] of 0.103, so this data can help us to calculate α.
3/2α = 0.103
α = 0.103 . 2/3 ⇒ 0.0686
So, concentration in equilibrium are
NH₃ = 0.0733 - 0.0733 . 0.0686 = 0.0682
N₂ = 0.0686/2 = 0.0343
So this moles, are in a volume of 1L, so they are molar concentrations.
Let's make Kc expression:
Kc= [N₂] . [H₂]³ / [NH₃]²
Kc = 0.0343 . 0.103³ / 0.0682² = 8.05x10⁻³
Answer:
ΔG°rxn = +50.8 kJ/mol
Explanation:
It is possible to obtain ΔG°rxn of a reaction at certain temperature from ΔH°rxn and S°rxn, thus:
<em>ΔG°rxn = ΔH°rxn - T×S°rxn (1)</em>
In the reaction:
2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l)
ΔH°rxn = 3×ΔHfNO2 + ΔHfH2O - (2×ΔHfHNO3 + ΔHfNO)
ΔH°rxn = 3×33.2kJ/mol + (-285.8kJ/mol) - (2×-207.0kJ/mol + 91.3kJ/mol)}
ΔH°rxn = 136.5kJ/mol
And S°:
S°rxn = 3×S°NO2 + S°H2O - (2×S°HNO3 + S°NO)
ΔH°rxn = 3×0.2401kJ/molK + (0.0700kJ/molK) - (2×0.146kJ/molK + 0.2108kJ/molK)
ΔH°rxn = 0.2875kJ/molK
And replacing in (1) at 298K:
ΔG°rxn = 136.5kJ/mol - 298K×0.2875kJ/molK
<em>ΔG°rxn = +50.8 kJ/mol</em>
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Answer:
correct option: A. tetrahedral
Explanation:
when we talk about hydrogen bonding in water due to oxygen then need to check the number of lone pair available for hydrogen bonding because hydrogen bond form between lone pair of oxygen and hydrogen of other water molecule so there are two lone pair which forms hydrogen bond and there are also two other covalent bond with hydrogen of same molecule so total number is 4 it means they will form a tetrahedral geometry
