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Fiesta28 [93]
3 years ago
11

The following reaction is exothermic. C6H12O6(s)+6O2(g)⇌6CO2(g)+6H2O(g)C6H12O6(s)+6O2(g)⇌6CO2(g)+6H2O(g) Predict the effect (shi

ft right, shift left, or no effect) of increasing and decreasing the reaction temperature.
Chemistry
2 answers:
Zigmanuir [339]3 years ago
6 0

Answer:

According to Le Chatelier's principle, increasing the reaction temperature of an exothermic reaction causes a shift to the left and decreasing the reaction temperature causes a shift to the right.

Explanation:

C6H12O6(s) + 6O2(g) ⇌6CO2(g) + 6H2O(g)

We are told that the forward reaction is exothermic, meaning heat is removed from the reacting substance to the surroundings.

According to Le Chatelier's principle,

1. for an exothermic reaction, on increasing the temperature, there is a shift in equilibrium to the left and formation of the product is favoured.

2. if the temperature of the system is decreased, the equilibrium shifts to right and the formation of the reactants is favoured.

3. if the reaction temperature is kept constant, the system is at equilibrium and there is no shift to the right nor to the left.

MAVERICK [17]3 years ago
6 0

Answer:

Increasing temperature = balance will shift to the left

Decreasing temperature = balance will shift to the right

Explanation:

Step 1: Data given

The increase or decrease in temperature can have an influence on the position of the equilibrium.

If the temperature is increased, the system will ensure that less heat is released. So the balance will shift to the left.

When the temperature drops, however, the system will produce more heat: the balance will shift to the right.

Step 2: The balanced equation

C6H12O6(s) + 6O2(g) ⇌ 6CO2(g) + 6H2O(g)

This is an endothermic reaction

Step 3: Increasing the temperature

If the temperature were increased, the heat content of the system would increase.

In exothermic reactions, increase in temperature decreases the K value. This means less products will be formed. The balance will shift to the left.

Step 4: Decreasing the temperature

If the temperature were decreased, the heat content of the system would increase.

In exothermic reactions, decrease in temperature increases the K value. This means more products will be formed, less reactants. The balance will shift to the right.

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Un mol de amoniaco Tiene una masa molar de 17 g y ocupa un volumen de 22.4 l qué volumen ocupa el 50 g amoniaco en condiciones n
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Answer:

V  = 65.81 L

Explanation:

En este caso, debemos usar la expresión para los gases ideales, la cual es la siguiente:

PV = nRT  (1)

Donde:

P: Presion (atm)

V: Volumen (L)

n: moles

R: constante de gases (0.082 L atm / mol K)

T: Temperatura (K)

De ahí, despejando el volumen tenemos:

V = nRT / P   (2)

Sin embargo como estamos hablando de condiciones normales de temperatura y presión, significa que estamos trabajando a 0° C (o 273 K) y 1 atm de presión. Lo que debemos hacer primero, es calcular los moles que hay en 50 g de amoníaco, usando su masa molar de 17 g/mol:

n = 50 / 17 = 2.94 moles

Con estos moles, reemplazamos en la expresión (2) y calculamos el volumen:

V = 2.94 * 0.082 * 273 / 1

<h2>V = 65.81 L</h2>
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Atomic mass of nitrogen
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Determine how many liters of hydrogen adjusted to STP there are in a 50.0 liter steel cylinder if the pressure inside is 100.0 a
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Answer : The volume of hydrogen gas at STP is 4550 L.

Explanation :

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The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 100.0 atm

P_2 = final pressure of gas at STP = 1 atm

V_1 = initial volume of gas = 50.0 L

V_2 = final volume of gas at STP = ?

T_1 = initial temperature of gas = 27.0^oC=273+27.0=300K

T_2 = final temperature of gas at STP = 0^oC=273+0=273K

Now put all the given values in the above equation, we get:

\frac{100.0atm\times 50.0L}{300K}=\frac{1atm\times V_2}{273K}

V_2=4550L

Therefore, the volume of hydrogen gas at STP is 4550 L.

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