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bekas [8.4K]
3 years ago
14

Catalytic converters made of palladium (pd) reduce automobile pollution by catalyzing the reaction between unburned hydrocarbons

and oxygen. how does pd increase the rate of this reaction
Chemistry
1 answer:
Doss [256]3 years ago
3 0
Unburned hydrocarbon on reacting with oxygen undergoes combustion reaction. However, the activation energy of this reaction is significantly high. When a catalyst like Pd is added to the reaction system, it provides active sites for the reaction to occur. It acts are a heterogeneous catalyst. It is pertinent of note that catalyst is refereed as heterogeneous, when it exist in different phase as compared to reactant and products. In present case, reactants and products are in gas phase, while catalyst is in solid phase. Due to availability of larger surface area at active site of Pd, activation energy of reaction decreases and decrease in activation energy favors higher reaction rates. 
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In an electron dot diagram of propane (C3H8), how many double bonds are present?
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Answer:- none

Explanations:- For naming hydrocarbons we use the suffix -ane, -ene and -yne.

-ane is used when we have only single bonds between all carbons. -ene is used if there is any double bond between two carbons and the -yne is used if there is any triple bond between two carbons.

The given name of the compound is propane. It ends at -ane and so it's an alkane and must have single bonds between all the carbons it has. So, there are zero double bonds present in C_3H_8 .

This is also clear from the below lewis dot structure of the compound.


4 0
3 years ago
Read 2 more answers
At 150°C the decomposition of acetaldehyde CH3CHO to methane is a first order reaction. If the
Crank

The decomposition time : 7.69 min ≈ 7.7 min

<h3>Further explanation</h3>

Given

rate constant : 0.029/min

a concentration of  0.050 mol L  to a concentration of 0.040 mol L

Required

the decomposition time

Solution

The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time

For first-order reaction :

[A]=[Ao]e^(-kt)

or

ln[A]=-kt+ln(A0)

Input the value :

ln(0.040)=-(0.029)t+ln(0.050)

-3.219 = -0.029t -2.996

-0.223 =-0.029t

t=7.69 minutes

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