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-Dominant- [34]
2 years ago
13

Write the expression for the equilibrium constant Kp for the following reaction.Enclose pressures in parentheses and do NOT writ

e the chemical formula as a subscript. For example, enter (PNH3)2 as (P NH3)2 .If either the numerator or denominator is 1, please enter 1NH4HS(s) ↔ NH3(g) + H2S(g)
Chemistry
1 answer:
maxonik [38]2 years ago
5 0

<u>Answer:</u> The expression for K_p is written below.

<u>Explanation:</u>

Equilibrium constant in terms of partial pressure is defined as the ratio of partial pressures of the products and the reactants each raised to the power their stoichiometric ratios. It is expressed as K_p

For a general chemical reaction:

aA+bB\rightleftharpoons cC+dD

The expression for K_p is written as:

K_p=\frac{P_{C}^c\times P_{D}^d}{P_{A}^a\times P_{B}^b}

The partial pressure for solids and liquids are taken as 1.

For the given chemical equation:

NH_4HS(s)\rightleftharpoons NH_3(g)+H_2S(g)

The expression for K_p for the following equation is:

K_p=\frac{(P NH_3)\times (P H_2S)}{(P NH_4HS)}

The partial pressure of NH_4HS will be 1 because it is solid.

So, the expression for K_p now becomes:

K_p=\frac{(P NH_3)\times (P H_2S)}{1}

Hence, the expression for K_p is written above.

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A 0.500 g sample of C7H5N2O6 is burned in a calorimeter containing 600. g of water at 20.0∘C. If the heat capacity of the bomb c
Nata [24]

Answer:

22.7

Explanation:

First, find the energy released by the mass of the sample. The heat of combustion is the heat per mole of the fuel:

ΔHC=qrxnn

We can rearrange the equation to solve for qrxn, remembering to convert the mass of sample into moles:

qrxn=ΔHrxn×n=−3374 kJ/mol×(0.500 g×1 mol213.125 g)=−7.916 kJ=−7916 J

The heat released by the reaction must be equal to the sum of the heat absorbed by the water and the calorimeter itself:

qrxn=−(qwater+qbomb)

The heat absorbed by the water can be calculated using the specific heat of water:

qwater=mcΔT

The heat absorbed by the calorimeter can be calculated from the heat capacity of the calorimeter:

qbomb=CΔT

Combine both equations into the first equation and substitute the known values, with ΔT=Tfinal−20.0∘C:

−7916 J=−[(4.184 Jg ∘C)(600. g)(Tfinal–20.0∘C)+(420. J∘C)(Tfinal–20.0∘C)]

Distribute the terms of each multiplication and simplify:

−7916 J=−[(2510.4 J∘C×Tfinal)–(2510.4 J∘C×20.0∘C)+(420. J∘C×Tfinal)–(420. J∘C×20.0∘C)]=−[(2510.4 J∘C×Tfinal)–50208 J+(420. J∘C×Tfinal)–8400 J]

Add the like terms and simplify:

−7916 J=−2930.4 J∘C×Tfinal+58608 J

Finally, solve for Tfinal:

−66524 J=−2930.4 J∘C×Tfinal

Tfinal=22.701∘C

The answer should have three significant figures, so round to 22.7∘C.

8 0
2 years ago
A 100. mL sample of 0.200 M aqueous hydrochloric acid is added to 100. mL of 0.200 M aqueous ammonia in a calorimeter whose heat
Free_Kalibri [48]

Answer:

-154KJ/mol

Explanation:

mole of 100ml sample of 0.2M aqueous HCl = Molarity × volume in Liter

= 0.2 × 100 / 1000  ( 1L = 1000 ml) = 0.02 mol and 0.02 mole of HCl solution require 0.02 mole of ammonia according to the mole ratio in the balanced equation.

Heat loss by the reaction = heat gain by calorimeter = mcΔT + 480 J/K

where m is the mass of water = 100g + 100g = 200g since mass of 100ml of water = 100g and it is in both of them and specific heat capacity of water 4.184 J/gK

heat gain by calorimeter  = (4.184 × 200 + 480) × 2.34 = 3081.3 J

ΔH per mole = heat loss / number of mole = 3081.3 / 0.02 = 154065.6 = -154KJ/mol

8 0
2 years ago
Steam initially at 0.3 MPa, 2500 C is cooled at constant volume. (a) At what temperature will steam become saturated vapour? [12
Scilla [17]

Answer:

a. 123.9°C

b.

c.

Explanation:

Hello, I'm attaching a picture with the numerical development of this exercise.

a. Since the steam is overheated vapour, the specific volume is gotten from the corresponding table. Then, as it became a saturated vapour, we look for the interval in which the same volume of state 1 is, then we interpolate and get the temperature.

b. Now, at 80°C, since it is about a rigid tank (constant volume for every thermodynamic process), the specific volume of the mixture is 0.79645 m^3/kg as well, so the specific volume for the liquid and the vapour are taken into account to get the quality of 0.234.

c. Now,since this is an isocoric process, the heat transfer per kg of steam is computed as the difference in the internal energy, considering the initial condition (showed in a. part) and the final one computed here.

** The thermodynamic data were obtained from Cengel's thermodynamics book 7th edition.

Best regards.

7 0
2 years ago
Mario uses a hot plate to heat a beaker of 50mL of water. He used a thermometer to measure the temperature of the water. The wat
Alex777 [14]

Answer:

The boiling point decreases as the volume decreases.

Explanation:

The Temperature - Volume law otherwise called as Charles law is applied, which says that the volume of the given gas at constant pressure is directly proportional to the temperature measured in Kelvin. As the volume increases, the temperature also increases, if the volume decreases, then the temperature also decreases.

As per the Charles law, here the volume is decreased from 50 ml to 25 ml so the boiling point also decreases.

8 0
2 years ago
Read 2 more answers
Which aqueous solution would exhibit a lower freezing point, 0.35 m k2so4 or 0.50 m kcl? explain and show necessary calculations
Umnica [9.8K]
Answer is: a lower freezing point has solution of K₂SO₄.

Change in freezing point from pure solvent to solution: ΔT =i · Kf · b.<span>
Kf - molal freezing-point depression constant for water is 1.86°C/m.
b -  molality, moles of solute per kilogram of solvent.
i - </span>Van't Hoff factor.<span>
b(K</span>₂SO₄<span>) = 0.35 m.
</span>b(KCl) = 0.5 m.
i(K₂SO₄) = 3.
i(KCl) = 2.
ΔT(K₂SO₄) = 3 · 0.35 m · 1.86°C/m.
ΔT(K₂SO₄) = 1.953°C.
ΔT(KCl) = 2 · 0.5 m · 1.86°C/m.
ΔT(KCl) = 1.86°C.


8 0
2 years ago
Read 2 more answers
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