Question

As ${\mathrm{NH}}_3(g)$ is decomposed into nitrogen gas and hydrogen gas at constant pressure and temperature, the volume of the product gases collected is twice the volume of ${\mathrm{NH}}_3$ reacted. Explain. As ${\mathrm{NH}}_3(g)$ is decomposed into nitrogen gas and hydrogen gas at constant volume and temperature, the total pressure increases by some factor. Why the increase in pressure and by what factor does the total pressure increase when reactants are completely converted into products? How do the partial pressures of the product gases compare to each other and to the initial pressure of ${\mathrm{NH}}_3$ ?

Short answer

In the decomposition of ammonia ($N{H}_3$) into nitrogen gas ($N_2$) and hydrogen gas ($H_2$), the volume of product gases is twice the volume of ammonia at constant pressure and temperature because the total number of moles of gas doubles. At constant volume and temperature, the total pressure increases because the number of moles of gas in the system also doubles, resulting in a factor by which the total pressure doubles. The final partial pressure of nitrogen gas is half the initial pressure of ammonia, while the final partial pressure of hydrogen gas is 1.5 times the initial pressure of ammonia.

Step-by-step solution

Write down the balanced chemical equation for the decomposition of ammonia

We need to write the balanced chemical equation for the decomposition of ammonia ($N{H}_3$) into nitrogen gas ($N_2$) and hydrogen gas ($H_2$). The balanced equation is: $$2N{H}_3(g)\to N_2(g)+3H_2(g)$$

Explain the change in volume at constant pressure and temperature

According to the balanced chemical equation, two moles of ammonia produce one mole of nitrogen gas and three moles of hydrogen gas, resulting in a total of four moles of product gases for every two moles of ammonia. Since the volume of a gas is proportional to its number of moles at constant pressure and temperature, the volume of the product gases (__$N_2$ + 3_$H_2$__) is twice the volume of the ammonia reacted for every two moles of ammonia (one mole of nitrogen and three moles of hydrogen are produced - four moles in total).

Explain the increase in pressure at constant volume and temperature

At constant volume and temperature, the increase in the number of moles of gas will lead to an increase in pressure. For this reaction, two moles of ammonia are converted into four moles of product gases. Therefore, the total number of moles of gas in the system doubles, resulting in an increase in pressure.

Calculate the factor by which the total pressure increases

At constant volume and temperature, the pressure of a gas is directly proportional to the number of moles. As two moles of ammonia are converted into four moles of product gases, the total pressure doubles when the reactants are completely converted into products.

Compare the partial pressures of the product gases and the initial pressure of ammonia

According to Dalton's law of partial pressures, the total pressure is the sum of the partial pressures of all gases present in the system. Since the total pressure doubles, we can deduce the partial pressures of the product gases as follows: 1. Initial pressure of ammonia: $P_{N{H}_3}^{initial}=P$ 2. Final pressure of nitrogen gas: Since one mole of nitrogen gas is produced for every two moles of ammonia, the partial pressure of nitrogen gas will be: $P_{N_2}^{final}=\frac{1}{4}\times 2P=\frac{1}{2}P$ 3. Final pressure of hydrogen gas: Since three moles of hydrogen gas are produced for every two moles of ammonia, the partial pressure of hydrogen gas will be: $P_{H_2}^{final}=\frac{3}{4}\times 2P=\frac{3}{2}P$ Thus, the final partial pressure of nitrogen gas is half the initial pressure of ammonia, while the final partial pressure of hydrogen gas is 1.5 times the initial pressure of ammonia.