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3 years ago

What will happen when additional reactants are added to a system at equilibrium ?

Chemistry

2 answers:

Pachacha [2.7K]3 years ago

5 0

Additional reactants would break equilibrium

Advocard [28]3 years ago

5 0

Answer:

The position of equilibrium will shift to the right.

Explanation:

Reactants ⇌ Products

According to Le Châtelier's Principle, when a stress is applied to a system at equilibrium, the system will respond in a way that tends to relieve the stress.

If we add additional reactants, the position of equilibrium (POE) will move to the right to get rid of the added reactants.

The concentration of products will increase.

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Need help !!!!! ASAP

Ksivusya [100]

<h2>Hello!</h2>

The answer is:

We have that there were produced 0.120 moles of $CO_{2}$

$n=0.120mol$

We are asked to calculate the number of moles of the given gas, also, we are given the volume, the temperature and the pressure of the gas, we can calculate the approximate volume using The Ideal Gas Law.

The Ideal Gas Law is based on Boyle's Law, Gay-Lussac's Law, Charles's Law, and Avogadro's Law, and it's described by the following equation:

$PV=nRT$

Where,

P is the pressure of the gas.

V is the volume of the gas.

n is the number of moles of the gas.

T is the absolute temperature of the gas (Kelvin).

R is the ideal gas constant (to work with pressure in mmHg), which is equal to:

$R=62.363\frac{mmHg.L}{mol.K}$

We must remember that the The Ideal Gas Law equation works with absolute temperatures (K), so, if we are given relative temperatures such as Celsius degrees or Fahrenheit degrees, we need to convert it to Kelvin before we proceed to work with the equation.

We can convert from Celsius degrees to Kelvin using the following formula:

$Temperature(K)=Temperature(C\°) + 273K$

So, we are given the following information:

$Pressure=760mmHg\\Volume=2.965L\\Temperature=25.5C\°=25.5+273K=298.5K$

Now, isolating the number of moles, and substituting the given information, we have:

$PV=nRT$

$n=\frac{PV}{RT}$

$n=\frac{760mmHg*2.965L}{62.363\frac{mmHg.L}{mol.K}*298.5K}$

$n=\frac{760mmHg*2.965L}{62.363\frac{mmHg.L}{mol.K}*298.5K}\\\\n=\frac{2242mmHg.L}{18615.355\frac{mmHg.L}{mol.}}\\\\n=0.120mole$

Hence, we have that there were produced 0.120 moles of $CO_{2}$

$n=0.120mol$

Have a nice day!

7 0

3 years ago

QUICK HELP PLEASE!!?? Need answer QUICK

d1i1m1o1n [39]

Answer:

Explanation:

Look on the x-axis for the tick marked "60". This indicates 60 degrees Celsius, which we want. Now, look on the y-axis for the tick marked "60". This indicates 60 grams of $Na_2HAsO_4$ . Trace along the graph to find where these two places meet at (60, 60).

Now, look for the solubility curve of $Na_2HAsO_4$ ; it's the yellow-orange line. Find out what the y-coordinate of the point where x = 60 is on the line: it's around (60, 65).

So, since the point (60, 60) is below the line corresponding to this substance, $Na_2HAsO_4$ is unsaturated.

The answer is B.

4 0

3 years ago

Read 2 more answers

Heat transfer putting a heating pad on leg

beks73 [17]

ANSWER IS CONDUCTION. HOPE THIS HELPED!

8 0

3 years ago

Be sure to answer all parts. for each reaction, find the value of δso. report the value with the appropriate sign. (a) 3 no2(g)

Maurinko [17]

Answer:

Change in entropy for the reaction is

ΔS° = -268.13 J/K.mol

Explanation:

To calculate the change in entropy for the balanced reaction, we require the natural entropy of all the reactants and products in the reaction.

3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)

From Literature.

S°(NO₂) = 240.06 J/K.mol

S°(H₂O) = 69.91 J/K.mol

S°(HNO₃) = 155.60 J/K.mol

S°(NO) = 210.76 J/K.mol

These are the entropies of the reactants and products under standard conditions of 298.15 K and 1 atm.

Note that

ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)

Σ nᵢS°(for products) = [2 × S°(HNO₃)] + [1 × S°(NO)]

= (2 × 155.60) + (1 × 210.76) = 521.96 J/K.mol

Σ nᵢS°(for reactants) = [3 × S°(NO₂)] + [1 × S°(H₂O)]

= (3 × 240.06) + (1 × 69.91) =790.09 J/K.mol

ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)

ΔS° = 521.96 - 790.09 = -268.13 J/K.mol

Hope this Helps!!

4 0

3 years ago

Give a brief description of the symptoms associated with dissociative fugue.

fomenos

Answer:

People may seem and act normally during the fugue, or they may appear moderately bewildered and draw no notice. When the fugue is over, however, people are thrown into a new scenario with no recall of how they got there or what they were doing.

Explanation:

3 0

2 years ago