Chemistry, 24.03.2020 17:01 emopandabogard8712

Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial chemist studying this reaction fills a tank with of sulfur dioxide gas and of oxygen gas, and when the mixture has come to equilibrium measures the amount of sulfur trioxide gas to be . Calculate the concentration equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture. Round your answer to significant digits.

Answers: 2

Show answers

Answers

Answer from: alexisbcatlett14

The question is incomplete, here is the complete question:

Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial chemist studying this reaction fills 125 L a tank with 60 moles of sulfur dioxide gas and 12 moles of oxygen gas at 30°C and when the mixture has come to equilibrium measures the amount of sulfur trioxide gas to be 12 moles.

Calculate the concentration equilibrium constant for the reaction of sulfur dioxide and oxygen at the final temperature of the mixture. Round your answer to 2 significant digits.

The concentration equilibrium constant for the reaction is 1.302

Explanation:

We are given:

Initial moles of sulfur dioxide = 60 moles

Initial moles of oxygen gas = 12 moles

Equilibrium moles of sulfur trioxide = 12 moles

Volume of tank = 125 L

Molarity is calculated by using the equation, we get:

$\text{Initial molarity of sulfur dioxide}=\frac{60}{125}=0.48M$

$\text{Initial molarity of oxygen gas}=\frac{12}{125}=0.096M$

$\text{Equilibrium molarity of sulfur trioxide}=\frac{12}{125}=0.096M$

The chemical equation for the reaction of sulfur dioxide and oxygen gas follows:

$2SO_2+O_2\rightleftharpoons 2SO_3$

Initial: 0.48 0.096

At eqllm: 0.48-2x 0.096-x 2x

Evaluating the value of 'x':

$\Rightarrow 2x=0.096\\\\x=0.048$

So, equilibrium concentration of sulfur dioxide = (0.48 - 2x) = [0.48 - 2(0.048)] = 0.384 M

Equilibrium concentration of oxygen = (0.096 - x) = [0.096 - 0.048] = 0.048 M

The expression of K_c for above equation follows:

$K_c=\frac{[SO_3]^2}{[O_2]\times [SO_2]^2}$

Putting values in above equation, we get:

$K_c=\frac{(0.096)^2}{0.048\times (0.384)^2}\\\\K_c=1.302$

Hence, the concentration equilibrium constant for the reaction is 1.302

Answer from: Quest

Since there is the same amount of agn03 in both silver chloride and silver nitrate they produce the same amount of moles. therefore silver chloride produces the same amount of silver nitrate, so your answer is 15.0 moles of silver nitrate. i hope this !

Answer from: Quest

answer is: nuclear fusion produces elements that are heavier than helium.

nuclei combine to form a heavier nucleus, releasing energy.

for example, fusion of hydrogen atoms at the sun.

this fusion nuclear reaction is in the inner core of the sun and hydrogen (atomic number 1) is conveted into helium (atomic number 2).

the inner core is the central region of the sun and all solar energy is produced in the core by nuclear fusion.