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prisoha [69]
2 years ago
9

For the reaction: 2 A (g) + B (s)= 2 C(s) + D (g)

Chemistry
1 answer:
jolli1 [7]2 years ago
8 0

Answer:

The value of the equilibrium constant = 5.213

Explanation:

Here K_p (equilibrium constant) is referred to as the  partial pressure of product divided by the  partial pressure of reactant with each pressure term raised to power that is equal to its stoichiometric coefficient in balanced equation .

As such only gas appear in K_p  expression as solids takes a value of 1;

SO ; in the  given equation from the question:

2 A (g) + B (s) ----> 2 C(s) + D (g)

K_p = \dfrac{[D]}{[A]^2}

K_p = \dfrac{2.71}{0.721^2}

K_p = 5.213

The value of the equilibrium constant = 5.213

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A bomb calorimetric experiment was run to determine the enthalpy of combustion of ethanol. The reaction is The bomb had a heat c
gulaghasi [49]

Answer:

The enthalpy of combustion of ethanol in kJ/mol is -1419.58  kJ/mol.

Explanation:

The heat absorbed by the bomb and water is equal to the product of the heat  capacity and the temperature change. Working with this equation, and assuming no heat is lost to  the surroundings, we write :

qcal= Ccal × ΔT= 490 J/K × 276.7 K= <u>135,583 J</u> = 135.58 kJ

Note we expressed the temperature change in K, because the heat capacity is written in K.

<u> Now that we have the heat of combustion, we need to calculate the molar heat.  </u>

Because qsystem = qrxn + qcal and qrxn = -qcal, the heat change of the reaction is -135.58 kJ.

This is the heat released by the combustion of 4.40 g of ethanol ; therefore, we can write  the <u>conversion factor as 135.58 kJ/ 4.40 g</u>.

The molar mass of ethanol is 46.07 g, so the heat of combustion of 1 mole of ethanol is :

molar heat of combustion= -135.58 kJ/4.40 g x 46.07 g/ 1 mol= -1419.58 kJ/mol

Therefore, the enthalpy of combustion of ethanol in kJ/mol is -1419.58  kJ/mol.

3 0
3 years ago
When 3.0 kg of water is warmed from 10 °C to 80 °C, how much heat energy is needed?
Naddik [55]

Answer:

THE HEAT NEEDED TO CHANGE 3KG OF WATER FROM 10 C TO 80 C IS 877.8kJ OR 877,800 J.

Explanation:

Mass = 3.0 kg = 3 * 1000 = 3000 g

Initial temperature = 10 C

Final temperature = 80 C

Change in temperature = 80 - 10 = 70 C

Specific heat of water = 4.18 J/g C

Heat needed = unknown

Heat is the amount of energy in joules needed to change a gram of water by 1 C.

Heat = mass * specific heat * change in temperature

Heat = 3000 g * 4.18 J/g C * 70 C

Heat = 877 800 Joules

Heat = 877.8 kJ.

The heat needed to change 3 kg mass of water from 10 C to 80 C is 877,800 J or 877.8 kJ.

6 0
3 years ago
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