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NCERT Exemplar · Q28

Q.45.4 L of dinitrogen reacted with 22.7 L of dioxygen and 45.4 L of nitrous oxide was formed. The reaction is given below: 2N2(g)+O2(g)→2N2O(g)2N_2(g) + O_2(g) \rightarrow 2N_2O(g) Which law is being obeyed in this experiment? Write the statement of the law?

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The experiment demonstrates Gay-Lussac's Law of Gaseous Volumes, as the reacting volumes of gases and the volume of gaseous product bear a simple whole-number ratio (2:1:22:1:2) at constant temperature and pressure.

When chemical reactions involve gases, their volumes often behave in a very specific and predictable way, provided the temperature and pressure remain constant. This predictability is a cornerstone of early chemical understanding and helps us relate macroscopic observations (volumes) to microscopic stoichiometry (moles). The key idea here is that for gases, under identical conditions, the volume occupied is directly proportional to the number of moles.

Understanding the Law of Gaseous Volumes

Imagine you have a balanced chemical equation. The coefficients in this equation tell you the molar ratios of reactants and products. For example, in the given reaction 2N2(g)+O2(g)→2N2O(g)2N_2(g) + O_2(g) \rightarrow 2N_2O(g), it means 2 moles of dinitrogen react with 1 mole of dioxygen to produce 2 moles of nitrous oxide.

Now, if all these substances are gases and are measured at the same temperature and pressure, a remarkable simplification occurs. According to Avogadro's Law, equal volumes of all gases, at the same temperature and pressure, contain an equal number of molecules (or moles). This means that the ratio of moles is directly equivalent to the ratio of volumes for gases under these conditions.

Therefore, the stoichiometric coefficients in a balanced chemical equation for gaseous reactions directly represent the ratios of their reacting and product volumes. This observation was first made by Joseph Louis Gay-Lussac.

Step-by-Step Solution

  1. Identify the balanced chemical equation and its stoichiometric ratios:

    The given reaction is:

    2N2(g)+O2(g)→2N2O(g)2N_2(g) + O_2(g) \rightarrow 2N_2O(g)

    From this equation, we can see the molar ratios of the reacting gases and the gaseous product:

    N2:O2:N2O=2:1:2N_2 : O_2 : N_2O = 2 : 1 : 2

  2. Recall Gay-Lussac's Law of Gaseous Volumes:

    This law states that when gases react together, they do so in volumes which bear a simple whole-number ratio to one another and to the volumes of the gaseous products, provided that all volumes are measured at the same temperature and pressure.

    For gaseous reactions at constant temperature and pressure, the ratio of volumes of reactants and products is equal to the ratio of their stoichiometric coefficients in the balanced chemical equation.

  3. Compare the experimental volumes with the stoichiometric ratios:

    The experimental volumes provided are:

    Volume of dinitrogen (N2N_2) reacted =45.4 L= 45.4 \text{ L}

    Volume of dioxygen (O2O_2) reacted =22.7 L= 22.7 \text{ L}

    Volume of nitrous oxide (N2ON_2O) formed =45.4 L= 45.4 \text{ L}

    Let's find the ratio of these experimental volumes: …

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