A certain catalyzed reaction is known to have an activation energy Ea 78.0 kJ/mo

Question

A certain catalyzed reaction is known to have an activation energy Ea 78.0 kJ/mol. Furthermore, the rate of this reaction is measured at 277. K and found to be 2.0x 10 Mis. Use this information to answer the questions in the table below. (You may need to use the scrollbar to see all the choices.) Suppose the concentrations of all reactants is kept the same, but the temperature is lowered by 5% from 277. K to 263 . K. | The rate will | V choose one stay the same rise about 5% rise more than 5% rise less than 5% fall about 5% fall more than 5% fall less than 5% How will the rate of the reaction change? Suppose the concentrations of all reactants is kept the same, but the catalyst is removed, which has the effect of raising the activation energy by 5%, from 78.0 kJ/mol to 81.9 kJ/mol The rate will How will the rate of the reaction change?

Explanation / Answer

Arhenius equation will be used to understand the effect of temperature and Activation energy. The equation can be written as

K= Ko*e(-Ea/RT)

K is rate constant, Ko is frequency factor, E is activation energy, T is temperature in K and R= gas constant= 8.314 J/mole.K

at two different conditions of temperature T1, T2 where the rate constants are K1 and K2, the equation becomes

ln(K2/K1)= (E/R)*(1/T1-1/T2)

given E= 78 Kj/mole= 78*1000J/mole, T1=277K, T2= 263 K and K1=2*10-6 M/s, K2=?

hecne ln(K2/K1)= (78*1000/8.314)*(1/277-1/263)

K2=3.3*10-7 M/s

% drop in rate constant = 100* {3.3*10-7-2*10-6)/(2*10-6)}=-83%, since there is a decrease in rate constant , the rate constant falls and it falls more than 5%.

when activation energy is increased, the rate constant decreases. At the same temperarure with different activation energies E2 and E1 and rate constants K2 and K1

the equation becomes ln (K2/K1)= (1/RT)*(E1-E2)

K2= rate constant in the absence of enzyme and K1= Rate constant in the presence of enznyme =2*10-6M/s

E1=activation energy in the presence of enzyme= 78 Kj/mole and E2= activation energy in the absence of enzyme= 81.9 Kj/mole

hence ln (K2/K1)= (1/8.314*277)*(78-81.9)*1000=3.7*10-7 M/s

so the rate constant decreases. The % decrease in rate constant is100*{ ( 3.7*10-7-2.6*10-6)/2.6*10-6}=-81.6% , there is fall in rate constant by more than 5%.

2. Total pressure = partial pressure of air + vapor pressure of ethanol.

As the temperatrue is increased, vapor pressure of ethanol increases and this also increases the total pressure and this also increases the rate of movement of gas molecules

3. Volatile chemicals will have more vapor pressure, nornally toxic and they cannot be heated in open containers since the vapors can form explosive mixture with air. Since the vapors are toxic, protective equipment is a must. Tightly closed containers prevent leakage of vapors. They should not be mixed with other chemicals. Except for second statement, rest statements are correct.

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