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2 years ago

What stress will shift the following equilibrium system to the left? 2SO2(g) + O2(g) ⇌ 2SO3(g); ΔH= –98.8 kJ/mol

Chemistry

1 answer:

svlad2 [7]2 years ago

6 0

Answer:

The stress that will shift the equilibrium system to the right is decreasing the temperature

Explanation:

Chemical equilibrium

This is a state where there is no observable changes in the properties of the system with time.

From the above, a French scientist establishes a principle which states that when an external constraints such as change in temperature, pressure and concentration is imposed on a system in equilibrium, the equilibrium will shift to neutralise the effect.

Furthermore, the principle explained that:

An increase in temperature of an exothermic reaction (i.e ΔH= –ve) will shift the equilibrium backward.

A decrease in temperature of an exothermic reaction (i.e ΔH= –ve) will shift the equilibrium forward.

An increase in temperature of an endothermic reaction (i.e ΔH= +ve) will shift the equilibrium forward.

A decrease in temperature of an exothermic reaction (i.e ΔH= +ve) will shift the equilibrium back.

How to determine which stress that will shift the equilibrium forward (right)

2SO₂(g) + O₂(g) ⇌ 2SO₃(g); ΔH= –98.8 KJ/mol

From the equation above, we can see that the reaction is exothermic reaction (i.e ΔH= –ve).

Therefore, decreasing the temperature will shift the equilibrium to the right.

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What mass of radon is in 8.17 moles of radon?

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Answer:

1813.74g

Explanation:

Given parameters:

Number of moles of radon = 8.17moles

Unknown:

Mass of radon = ?

Solution:

To solve this problem, we use the expression below:

Number of moles = $\frac{mass}{molar mass}$

Molar mass of radon = 222g/mol

Now insert the parameters and solve;

Mass of radon = Number of moles x molar mass

= 8.17 x 222

= 1813.74g

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3 years ago

When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H

Debora [2.8K]

Answer:

$Y=58.15\%$

Explanation:

Hello,

For the given chemical reaction:

$Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)$

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

$n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3$

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

$n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol Al_2S_3$

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

$m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3$

Finally, we compute the percent yield with the obtained 2.10 g:

$Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%$

Best regards.

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3 years ago