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ruslelena [56]
2 years ago
10

Using the equation 40kJ+2SO3(g)-->2SO2(g) + O2(g), if pressure is added what way

Chemistry
1 answer:
Ad libitum [116K]2 years ago
6 0

In an equation, 40kJ + 2SO₃ (g) -->2SO₂(g) + O₂(g), if pressure added, the equilibrium shifts to the left side of the reaction because it has fewer moles of gas.

<h3>What is Le Ch atelier's principle?</h3>

Le Ch atelier's principle states that if the pressure is added in a nay reaction, then the equilibrium will be shifted toward the side where there are fewer atoms.

In this case, there are fewer moles on the left side.

Thus, when pressure is added, the equilibrium shift to the left side due to Le Ch atelier's principle.

Learn more about Le Ch atelier's principle

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The concentration of Rn−222 in the basement of a house is 1.45 × 10−6 mol/L. Assume the air remains static and calculate the con
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<u>Explanation:</u>

All the radioactive reactions follows first order kinetics.

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t_{1/2}=\frac{0.693}{k}

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t_{1/2}=3.82days

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k=\frac{0.693}{3.82}=0.181days^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,  

k = rate constant = 0.181days^{-1}

t = time taken for decay process = 3.00 days

[A_o] = initial amount of the reactant = 1.45\times 10^{-6}mol/L

[A] = amount left after decay process =  ?

Putting values in above equation, we get:

0.181days^{-1}=\frac{2.303}{3.00days}\log\frac{1.45\times 10^{-6}}{[A]}

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Hence, the concentration of radon after the given time is 3.83\times 10^{-30}mol/L

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