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MAVERICK [17]
2 years ago
8

How do you think temperature will affect the rate of a chemical reaction?

Chemistry
1 answer:
Luba_88 [7]2 years ago
8 0

Answer:

I could create a slower reaction because the particles might be moving slower due to the cold. if it was warm there will be a faster reaction. similar to the elements movements in solids and liquids.

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The equilibrium constant for the dissolution of silver chloride (AgCl(s) Ag+(aq) + Cl–(aq)) has a value of 1.79 × 10–10. Which s
ryzh [129]

"Silver chloride is essentially insoluble in water" this statement is true for the equilibrium constant for the dissolution of silver chloride.

Option: b

<u>Explanation</u>:

As silver chloride is essentially insoluble in water but also show sparing solubility, its reason is explained through Fajan's rule. Therefore when AgCl added in water, equilibrium take place between undissolved and dissolved ions. While solubility product constant \left(\boldsymbol{K}_{s p}\right) for silver chloride is determined by equilibrium concentrations of dissolved ions. But solubility may vary also at different temperatures.  Complete solubility is possible in ammonia solution as it form stable complex as water is not good ligand for Ag+.  

To calculate \left(\boldsymbol{K}_{s p}\right) firstly molarity of ions are needed to be found with formula: \text { Molarity of ions }=\frac{\text { number of moles of solute }}{\text { Volume of solution in litres }}

Then at equilibrium cations and anions concentration is considered same hence:

\left[\mathbf{A} \mathbf{g}^{+}\right]=[\mathbf{C} \mathbf{I}]=\text { molarity of ions }

Hence from above data \left(\boldsymbol{K}_{s p}\right) can be calculated by: \left(\boldsymbol{K}_{s p}\right) = \left[\mathbf{A} \mathbf{g}^{+}\right] \cdot[\mathbf{C} \mathbf{I}]

6 0
3 years ago
Bromine reacts with nitric oxide to form nitrosyl bromide as shown in this reaction: br2(g) + 2 no(g) → 2 nobr(g) a possible mec
Svetach [21]

The overall reaction is given by:

Br_{2}(g) + 2 NO(g) \rightarrow  2 NOBr(g)

The fast step reaction is given as:

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The slow step reaction is given as:

NOBr_{2}(g) + NO(g) \rightarrow  2 NOBr(g) (slow step k_{2})

Now, the expression for the rate of reaction of fast reaction is:

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The expression for the rate of reaction of slow reaction is:

r_{2}=k_{2}[NOBr_{2}] [NO]

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The expression of rate of formation is:

\frac{d(NOBr)}{dt}=r_{2}

= k_{2}[NOBr_{2}][NO]    (1)

Now, consider that the fast step is always is in equilibrium. Therefore, r_{1}=0

k_{1}[NO][Br_{2}]= k_{-1}[NOBr_{2}]

[NOBr_{2}] = \frac{k_{1}}{k_{-1}}[NO][Br_{2}]

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=k_{2} \frac{k_{1}}{k_{-1}}[NO][Br_{2}][NO]

= \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]

Thus, rate law of formation of NOBr in terms of reactants is given by \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}].









4 0
3 years ago
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