03.03.2023 - 23:00

# Sulfuric acid is essential to dozens of important industries from steel making to plastics and pharmaceuticals. More sulfuric acid is made than any other industrial chemical, and world production exceeds 2.0 \times 10^{11} kg per year.The first step in th

Sulfuric acid is essential to dozens of important industries from steel making to plastics and pharmaceuticals. More sulfuric acid is made than any other industrial chemical, and world production exceeds {eq}2.0 \times 10^{11}\ \rm{kg} {/eq} per year.The first step in the synthesis of sulfuric acid is usually burning solid sulfur to make sulfur dioxide gas. Suppose an engineer studying this reaction introduces {eq}2.1\ \rm{kg} {/eq} of solid sulfur and {eq}5.80\ \rm{atm} {/eq} of oxygen gas at {eq}450^\circ \rm{C} {/eq} into an evacuated {eq}35.0\ \rm{L} {/eq} tank. The engineer believes {eq}K_p = 5.5 {/eq} for the reaction at this temperature. Calculate the mass of solid sulfur he expects to be consumed when the reaction reaches equilibrium. Round your answer to 2 significant digits.

(Note for advanced students: The engineer may be mistaken in his belief about the value of {eq}K_p {/eq}, and the consumption of sulfur you calculate may not be what he actually observes.)

Answers (1)
• April 8, 2023 в 05:36

The chemical equation for the reaction between solid sulfur and oxygen gas to form sulfur dioxide gas is:

{eq}\rm S(s) + O_2(g) rightarrow SO_2(g) {/eq}

The engineer introduced {eq}2.1; kg {/eq} of solid sulfur and {eq}5.80; atm {/eq} of oxygen gas at {eq}450^circ C {/eq} into a {eq}35.0; L {/eq} tank. The reaction takes place in an evacuated tank, meaning that the initial pressure of all gases is zero.

The first step in solving this problem is to calculate the number of moles of oxygen gas present in the tank:

{eq}\begin{aligned} n_{rm O_2} &= d\frac{PV}{RT} &= d\frac{(5.80; rm atm) \times (0.0350; rm m^3)}{(0.08206; rm Lcdot atmcdot K^{-1}cdot mol^{-1}) \times (723; rm K)} &= 0.914; rm mol \end{aligned} {/eq}

Next, we need to determine the limiting reactant in the reaction. To do this, we calculate the number of moles of sulfur that was introduced:

{eq}\begin{aligned} n_{rm S} &= d\frac{m_{rm S}}{M_{rm S}} &= d\frac{2.1; rm kg}{32.1; rm g/mol} &= 65.4; rm mol \end{aligned} {/eq}

The stoichiometry of the reaction tells us that 1 mole of sulfur reacts with 1 mole of oxygen to produce 1 mole of sulfur dioxide. Therefore, the maximum number of moles of sulfur dioxide that can be produced from the given amounts of sulfur and oxygen is:

{eq}n_{rm SO_2} = min(n_{rm S}, n_{rm O_2}) = 0.914; rm mol {/eq}

If the reaction were to go to completion, all of the oxygen gas would be consumed, and there would be excess sulfur left over. However, the problem asks us to find the mass of sulfur that is consumed when the reaction reaches equilibrium. This means that we need to calculate the equilibrium constant for the reaction and use it to determine the extent to which the reaction proceeds.

The equilibrium constant for the reaction is given by:

{eq}K_p = d\frac{P_{rm SO_2}}{P_{rm O_2}^{1/2}} {/eq}

where {eq}P_{rm SO_2} {/eq} and {eq}P_{rm O_2} {/eq} are the partial pressures of sulfur dioxide and oxygen, respectively, at equilibrium.

At the start of the reaction, the pressure of sulfur dioxide is zero, so we can assume that the reaction proceeds in the forward direction. As the reaction progresses, the pressure of sulfur dioxide increases, while the pressure of oxygen decreases. At some point, the forward and reverse rates of the reaction become equal, and the system reaches equilibrium.

Let {eq}x {/eq} be the number of moles of sulfur that reacts with oxygen to form sulfur dioxide at equilibrium. Then, the number of moles of oxygen and sulfur dioxide at equilibrium are both given by {eq}0.914 - x {/eq}. The pressure of oxygen at equilibrium is:

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