Gas Evolution Equations - Online Tutor, Practice Problems & Exam Prep

Understanding gas evolution reactions is crucial in chemistry, particularly when dealing with the formation of gases like ammonia (NH₃), carbon dioxide (CO₂), and sulfur dioxide (SO₂). These gases typically arise when intermediate products, which are not yet the final substances, lose a water molecule. For instance, when hydroxide ions react with ammonium ions, the intermediate product is NH₄OH, which upon losing water, forms NH₃, ammonia gas. Similarly, when hydrogen ions react with bicarbonate ions, carbonic acid (H₂CO₃) forms as an intermediate. The loss of water from carbonic acid yields CO₂. In the case of sulfurous acid, removing water results in SO₂. Notably, hydrogen sulfide (H₂S) is already in its final gas form and does not require the loss of water. These reactions are pivotal in predicting the final gaseous products in chemical equations.

Gas Evolution Equations

A gas evolution equation is a molecular equation that involves the creation of specific gases. Now the gases that we're talking about are NH3, which is ammonia, CO2, which is carbon dioxide, and SO2, which is sulfur dioxide. These gases are formed once medium products lose a water molecule. Now what exactly is a medium product? Well, a medium product is just the former product holds before it fully converts into its final product form. Now the final product form equals your medium product minus the water that we lose.

So here if we take a look, we have our reactant ions. So these ions will combine together to give me my medium product. So here we have hydroxide ion with your ammonium ion. Remember, when the numbers in the charges are the same, they just combine together and the charges cancel out. So here we're going to have as our medium product NH4OH. Next we have H+ and HCO3-, so hydrogen ion and bicarbonate ion. Again, when they combine together the charges cancel out, so H2CO3 which should be familiar as carbonic acid. Here the numbers in the charges are different. When they're different they don't cancel out, they crisscross. And when you do that, realize that we still make carbonic acid.

Next we have H+ and SO32-. Again, the numbers and the charges are different, so they don't cancel out, they crisscross. So here we make sulfurous acid and then finally we have H+ with S2-, our sulfide ion. The numbers in the charges are different, so they don't cancel out, they just crisscross. So here we're going to make H2S. Now how do we get to our final product? We say that we subtract water from each one of the medium products. So if I'm subtracting water from each of these, we see what's left behind. And what's left behind is NH3. So we lost H2O. Also removed one hydrogen here, this hydrogen here, and this oxygen. What's left is NH3. So that's how we get ammonia as our final gas product.

These two subtract H2O from them, so both hydrogen in front are gone. Remove one oxygen here because you're losing water and what's left is CO2, carbon dioxide. Here, sulfurous acid. Remove H2O. What's left is sulfur dioxide. Now, this last one here, this medium form actually isn't really a medium form. This is the final form of our gas, hydrogen sulfide gas. So here there is no oxygen to lose from it, so it's not losing water. So you create H2S initially, it stays H2S. So in a gas evolution equation, these are the gases that you tend to see as our final products. So just remember gas evolution, we create these products, at least one of them as our final product. So we start doing example questions and practice questions. We'll see how this comes about.

Gas Evolution Equations Example 1

Here in this example question, it says predict whether a reaction occurs and write the balanced molecular equation O. Here we have sodium carbonate reacting with hydrobromic acid. All right, so step one, we're going to break up reaction one and reactant 2 into their ionic forms. Sodium carbonate is composed of sodium ion. It's in Group 1A, so it's plus one, and carbonate ion which is a polyatomic ion, so it's CO32- hydrobromic acid is composed of H ion H1 Bromine is in Group 7A, so it's -1.

Step two, we swap ionic partners by remembering that opposite charges attract. So we're going to apply the rules that we've learned when it comes to creating new compounds from their ionic forms. If we take a look, this positive and this negative are attracted to one another, so they're going to combine. And remember, when your numbers and your charges are the same, the two ions just combine together and the charges disappear. So this will give us NaBr plus then we have here the numbers and the charges are different. When they're different, they don't just disappear and combine, they crisscross. The two from here comes here and the one from here comes here. That would give us, at this point H2CO3.

Now remember our solubility rules. NaBr Na is a group 1A ion. Anything connected to a group 1A ion is aqueous and soluble. And then look, we made H2CO3 that is one of our median products. So step three says identify the median product or gas that forms from the gas evolution equation except for hydrogen sulfide. Break it up into water and gas. So here we have carbonic acids which we know will break down further into water which will be a liquid plus CO2 gas South. You do you do not keep the carbonic acid there, it completely breaks down to give me those two pieces.

Step four. Last step says if necessary balance your molecular equation by placing the correct coefficients in front of each molecule. So here we just have to see is it balanced? We have here 2 sodiums and here we only have one, so I'd put a two in front of this CO3. So we have one carbon and three oxygens. We have one carbon 2 oxygens, 3 oxygens. So there goes the CO3 there hydrogens. We have one hydrogen here and one BR here, but on the product side we have two BRS because the two gets distributed to the BR and two hydrogens. So we'd have to put a 2 here. Now our equation is balanced, so these would be all the coefficients for this balanced gas evolution equation.

Predict the products formed from the following gas evolution equation.

CaS (aq) + HNO₃ (aq) →

• A. CaS (aq) + 2 HNO₃ (aq) → Ca(NO₃)₂ (aq) + 2 H₂S (g)
• B. CaS (aq) + HNO₃ (aq) → CaNO₃ (aq) + H₂S (g)
• C. CaS (aq) + 2 HNO₃ (aq) → Ca(NO₃)₂ (aq) + H₂S (g)
• D. No reaction

Predict the products formed from the following gas evolution equation.

NH₄Cl (aq) + NaOH (aq) →

• A. NH₄Cl (aq) + NaOH (aq) → NH₄OH (aq) + NaCl (aq)
• B. NH₄Cl (aq) + NaOH (aq) → NH₃ (g) + H₂O (l) + NaCl (aq)
• C. NH₄Cl (aq) + NaOH (aq) → NH₄OH (g) + NaCl (s)
• D. No reaction

Predict the products formed from the following gas evolution equation

K₂SO₃ (aq) + H₂SO₄ (aq) →

• A. K₂SO₃ (aq) + H₂SO₄ (aq) → K₂SO₄ (aq) + H₂O (l) + SO₂ (g)
• B. K₂SO₃ (aq) + H₂SO₄ (aq) → K₂SO₄ (aq) + H₂SO₃ (aq)
• C. K₂SO₃ (aq) + H₂SO₄ (aq) → K₂SO₄ (aq) + SO₂ (g)
• D. No reaction

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What is a gas evolution reaction?

A gas evolution reaction is a type of chemical reaction where the reactants interact to form one or more gaseous products. This process often results in the release of a gas, which can be observed as bubbles or effervescence in a liquid reaction mixture. Gas evolution reactions are common in various fields of chemistry, including inorganic and organic chemistry, and can be involved in processes such as photosynthesis, respiration, and industrial chemical production.

For example, when you mix hydrochloric acid (HCl) with sodium bicarbonate (NaHCO₃), you get sodium chloride (NaCl), water (H₂O), and carbon dioxide gas (CO₂) as products. The carbon dioxide gas forms bubbles and escapes from the mixture, which is a classic example of a gas evolution reaction:

${\mathrm{HCl}}_{\text{(aq)}}+{\mathrm{NaHCO}}_3(s)\; \to {\mathrm{NaCl}}_{\text{(aq)}}+H_{2}O(l)\; +{\mathrm{CO}}_2(g)$

Gas evolution reactions are important to identify because the gases produced can sometimes be hazardous, and knowing when a gas is going to be released can be crucial for conducting safe experiments or industrial processes.