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Lerok [7]
3 years ago
13

As shown in table 15.2, kp for the equilibrium n21g2 + 3 h21g2 δ 2 nh31g2 is 4.51 * 10-5 at 450 °c. for each of the mixtures lis

ted here, indicate whether the mixture is at equilibrium at 450 °c. if it is not at equilibrium, indicate the direction (toward product or toward reactants) in which the mixture must shift to achieve equilibrium. (a) 98 atm nh3, 45 atm n2, 55 atm h2 (b) 57 atm nh3, 143 atm n2, no h2 (c) 13 atm nh3, 27 atm n2, 82 atm h2
Chemistry
1 answer:
Setler79 [48]3 years ago
3 0
Our reaction balanced equation at equilibrium N2(g) + 3 H2(g) ↔ 2 NH3(g)
and we have the Kp value at equilibrium = 4.51 X 10^-5
A) 98 atm NH3, 45 atm N2, 55 atm H2

when Kp = [P(NH3)]^2 / [P(N2)] * [P(H2)]^3
                = 98^2 / (45 * 55^3) = 1.28 x 10^-3
by comparing the Kp by the Kp at equilibrium(the given value) So,
Kp > Kp equ So the mixture is not equilibrium,
 it will shift leftward (to decrease its value) towards the reactants to achieve equilibrium.
B) 57 atm NH3, 143 atm N2, no H2
∴ Kp = [P(NH3)]^2 / [P(N2)]
         = 57^2 / 143 = 22.7
∴Kp> Kp equ (the given value) 
∴it will shift leftward (to decrease its value) towards reactants to achieve equilibrium.

c) 13 atm NH3, 27 atm N2, 82 atm H2
∴Kp = [P(NH3)]^2 / [P(N2)] * [P(H2)]^3
       =  13^2 / (27* 82^3) = 1.14 X 10^-5
∴ Kp< Kp equ (the given value)
∴it will shift rightward (to increase its value) towards porducts to achieve equilibrium.


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Answer:

P' = 41.4 mmHg → Vapor pressure of solution

Explanation:

ΔP = P° . Xm

ΔP = Vapor pressure of pure solvent (P°) - Vapor pressure of solution (P')

Xm = Mole fraction for solute (Moles of solvent /Total moles)

Firstly we determine the mole fraction of solute.

Moles of solute → Mass . 1 mol / molar mass

20.2 g . 1 mol / 342 g = 0.0590 mol

Moles of solvent → Mass . 1mol / molar mass

60.5 g . 1 mol/ 18 g = 3.36 mol

Total moles = 3.36 mol + 0.0590 mol = 3.419 moles

Xm = 0.0590 mol / 3.419 moles → 0.0172

Let's replace the data in the formula

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P' = - (42.2 mmHg . 0.0172 - 42.2 mmHg)

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3 years ago
If 3.00 mL of 0.0250 M CuSO4 is diluted to 25.0 mL with pure water, what is the molarity of copper(II) sulfate in the diluted so
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Answer:

0.00268 M

Explanation:

To find the new molarity, you need to (1) find the moles of CuSO₄ (via the molarity equation using the beginning molarity and volume) and then (2) find the new molarity (using the moles and combined volume). Your final answer should have 3 sig figs to match the given values.

<u>Step 1:</u>

3.00 mL / 1,000 = 0.00300 L

Molarity = moles / volume (L)

0.0250 M = moles / 0.00300 L

(0.0250 M) x (0.00300 L) = moles

7.50 x 10⁻⁵ = moles

<u>Step 2:</u>

25.0 mL / 1,000 = 0.0250 L

0.0250 L + 0.00300 L = 0.0280 L

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3 years ago
A student performs the reduction of 4-nitrobenzaldehyde (151.12 g/mol) with sodium borohydride (37.83 g/mol) in the presence of
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Answer:

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m = 1.034 g

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The mass of the product that was expected is then:

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