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If a mixture containing 3 moles of hydrogen and 1 mole of nitrogen is converted completely into ammonia, the ratio of volumes of reactants and products at the same temperature and pressure would be: (a) \(2: 1\) (b) \(1: 2\) (c) \(1: 3\) (d) \(3: 1\)

Short Answer

Expert verified
The ratio of volumes of reactants to products is \(2: 1\) (option a).

Step by step solution

01

Write the Balanced Equation

The chemical reaction for the formation of ammonia from hydrogen and nitrogen is given by: \[ N_2 + 3H_2 \rightarrow 2NH_3 \]. This balanced equation shows that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.
02

Analyze Moles of Reactants and Products

According to the balanced equation, 1 mole of \( N_2 \) and 3 moles of \( H_2 \) produce 2 moles of \( NH_3 \). Therefore, when all the reactants are used up, 4 moles of gas (1 from \( N_2 \) + 3 from \( H_2 \)) give 2 moles of gas (\( NH_3 \)).
03

Calculate Volume Ratio

Under the same conditions of temperature and pressure, the number of moles is directly proportional to volume (Avogadro's Law). Thus, 4 moles of reactants convert to 2 moles of products, resulting in a volume ratio of \(4:2\). Simplifying this ratio gives \(2:1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stoichiometry
Understanding stoichiometry is essential when dealing with chemical reactions, as it helps us calculate the proportion of each reactant and product. It encompasses the quantitative relationship between reactants and products in a chemical reaction. In our example, the stoichiometry is expressed through the balanced chemical equation for the production of ammonia:
  • The equation is: \[ N_2 + 3H_2 \rightarrow 2NH_3 \]
Here, the coefficients in the equation indicate the stoichiometric ratios. We see that 1 mole of nitrogen reacts with 3 moles of hydrogen, illustrating a 1:3 stoichiometric ratio between hydrogen and nitrogen. This ratio is crucial for predicting the amount of product formed when given quantities of reactants are used. Remember, stoichiometry helps us achieve precise and optimal uses of resources in chemical processes.
Avogadro's Law
Avogadro's Law is a fundamental principle in chemistry that connects gas volume with the amount of moles. According to this law, equal volumes of gases at the same temperature and pressure contain an equal number of moles. This law is helpful when comparing volumes in reactions involving gases. For our chemical reaction involving ammonia:
  • The total number of moles on the reactants side is 4 (1 mole of nitrogen and 3 moles of hydrogen).
  • On the product side, we have 2 moles of ammonia gas.
Hence, when 4 moles of gas react to produce 2 moles of product at constant temperature and pressure, the volume ratio follows Avogadro's Law, reflecting the mole ratio precisely as volume ratio.
Balanced Chemical Equation
A balanced chemical equation is foundational to understanding the relationship between reactants and products. It ensures the conservation of atoms, meaning the same number of each type of atom is present on both sides of the reaction. Let's break down the balanced equation of ammonia production:
  • In \[ N_2 + 3H_2 \rightarrow 2NH_3 \]
  • 1 nitrogen molecule and 3 hydrogen molecules react.
  • Yielding 2 ammonia molecules without any leftover atoms, showing full conversion.
Such balance is essential for correctly applying stoichiometry as it provides the exact ratios of reactants and products, allowing predictions of how much of each substance will be involved in the reaction. This helps in scaling reactions up or down based on need.
Gas Volume Ratios
Gas volume ratios offer insight into how gases behave during chemical reactions under constant conditions. Based on our balanced equation: \[ N_2 + 3H_2 \rightarrow 2NH_3 \], the reaction starts with 4 moles of gaseous reactants and ends with 2 moles of gaseous products.
  • This gives us a volume ratio of reactants to products as \[ 4:2 \].
  • When simplified, becomes \[ 2:1 \].
This simplification is possible because chemical reactions involving gases adhere to Avogadro's Law, where the volume directly corresponds to mole quantity if conditions are constant. So, the volume of reactants will shrink post-reaction, following the stoichiometric guidance of the balanced equation. Understanding these ratios is critical in chemical engineering and practical applications where gas management is necessary.

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Most popular questions from this chapter

The equilibrium between water and its vapour, in an open vessel: (a) Can be achieved (b) Depends upon pressure (c) Cannot be achieved (d) Depends upon temperature

In the reaction, \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}+\) heat, relationship between \(\mathrm{K}_{\mathrm{P}}\) and \(\mathrm{K}_{\mathrm{c}}\) is: (a) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{-2}\) (b) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{2}\) (c) \(K_{p}=K_{c}(R T)^{-3}\) (d) \(\mathrm{K}_{\mathrm{c}}=\mathrm{K}_{\mathrm{p}}(\mathrm{RT})^{3}\)

The equilibrium constant for the following reaction will be \(3 \mathrm{~A}+2 \mathrm{~B} \rightleftharpoons \mathrm{C}\) : (a) \(\frac{[3 \mathrm{~A}][2 \mathrm{~B}]}{[\mathrm{C}]}\) (b) \(\frac{[\mathrm{C}]}{[3 \mathrm{~A}][2 \mathrm{~B}]}\) (c) \(\frac{[\mathrm{C}]}{[\mathrm{A}]^{2}[\mathrm{~B}]^{2}}\) (d) \(\frac{[\mathrm{C}]}{[\mathrm{A}]^{3}[\mathrm{~B}]^{2}}\)

One of the following equilibria is not affected by change in volume of the flask: (a) \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g}) \mathrm{Cl}_{2}(\mathrm{~g})\) (b) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) (c) \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\) (d) \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\)

An amount of solid \(\mathrm{NH}_{4}\) HS in placed in a flask already containing ammonia gas at a certain temperature and \(0.50\) atm pressure. Ammonium hydrogen sulphide decomposes to yield \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{~S}\) gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure is the flask rises to \(0.84 \mathrm{~atm}\), the equilibrium constant for \(\mathrm{NH}_{4} \mathrm{HS}\) decomposition at this temperature is: (a) \(0.30\) (b) \(0.18\) (c) \(0.17\) (d) \(0.11\)

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