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Basile [38]
2 years ago
6

B. The highly toxic gas phosgene is produced by the reaction CO(g) + Cl2(g) COCl2(g), ΔH = –108 kJ/mol. Use this reaction to ans

wer the following questions.
i. How would increasing the pressure of the system affect the reaction yield? Explain your answer. (2 points)

















ii. How would increasing the pressure of the system affect the reaction rate? Explain your answer. (2 points)

















iii. How would increasing the temperature of the system affect the reaction yield? Explain your answer. (2 points)

















iv. How would increasing the temperature of the system affect the reaction rate? Explain your answer. (2 points)
Chemistry
1 answer:
galben [10]2 years ago
6 0

Answer:

I. Increasing pressure will allow more frequent successful collision between particles due to the particles being closer together.

II. Rate of reaction increases due to more products being made; as increased pressure favours the exothermic side of the equilibrium.

III. Increasing temperature provides particles lots of (Kinetic) energy, for more frequent successful collision due to the particles moving at a faster rate than before. However, favouring the endothermic side of the equilibrium due to lots of energy required to break and form new bonds.

IV. Rate of reaction increases due to increase temperature favouring both directions of the equilibrium - causing products to form faster.

Hope this helps!

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If the initial concentration of SO2Cl2 is 0.125 M , what is the concentration of SO2Cl2 after 210 s ?
andre [41]

Complete question:

The decomposition of SO2Cl2 is first order in SO2Cl2 and has a rate constant of 1.44×10⁻⁴ s⁻¹ at a certain temperature.

If the initial concentration of SO2Cl2 is 0.125 M , what is the concentration of SO2Cl2 after 210 s ?

Answer:

After 210 s the concentration of SO2Cl2  will be 0.121 M

Explanation:

ln\frac{[A_t]}{[A_0]} =-kt

where;

At is the concentration of A at a time t

A₀ is the initial concentration of A

k is rate constant = 1.44×10⁻⁴ s⁻¹

t is time

ln(At/A₀) = -( 1.44×10⁻⁴)t

ln(At/0.125) =  -( 1.44×10⁻⁴)210

ln(At/0.125) = -0.03024

\frac{A_t}{0.125} = e^{-0.03024

At/0.125 = 0.9702

At = 0.125*0.9702

At = 0.121 M

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