For the given reactions- following data is given P - Q k1 - 1015 exp -2000-T C - D k2 - 1014 exp -1000-T Temperature at which k1 - k2 is-

The correct option is A 434.22 KAs we know, k_1 = 10^15 exp ( 2000/T ); k_2 = 10^14 exp ( 1000/T ) When k_1 = k_2; nbsp;10^15 exp ( 200 ...

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Standard XII

Chemistry

Arrhenius Equation

For the given...

Question

For the given reactions, following data is given
$P \to Q k_{1} = 10^{15} e x p (\frac{- 2000}{T})$
$C \to D k_{2} = 10^{14} e x p (\frac{- 1000}{T})$
Temperature at which $k_{1} = k_{2}$ is:

434.22 K

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1000 K

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2000 K

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868.44 K

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Solution

The correct option is A 434.22 K
As we know,
$k_{1} = 10^{15} e x p (\frac{- 2000}{T}); k_{2} = 10^{14} e x p (\frac{- 1000}{T})$
When $k_{1} = k_{2}; 10^{15} e x p (\frac{- 2000}{T}) = 10^{14} e x p (\frac{- 1000}{T})$
Or T = 434.22 K

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Similar questions

Q. For gaseous reaction, following data is given:

A \to B, k_{1} = 10^{15} e^{- 2000 / T}

C \to D, k_{2} = 10^{14} e^{- 1000 / T}

The temperature at which

k_{1} = k_{2}

is:

The rate of reaction increase significantly with increase in temperature. Generally, rates of reactions are doubled for every

10^{\circ}

rise in temperature. Temperature coefficient gives us an idea about the change in the rate of a reaction for every

10^{\circ}

change in temperature.
Temperature coefficient

(μ) = \frac{R a t e c o n s t a n t o f (T + 10)^{\circ} C}{R a t e c o n s t a n t a t T^{\circ}}

Arrhenius gave an equation which describes rate constant k as a function of temperature

k = A e^{- E_{a} / R T}

where k is the rate constant, A is the frequency factor or pre-exponetial factor,

E_{a}

is the activation energy, T is the teperature in kelvin, and R is the universal gas constant.
Equation when expressed in logarithmic form becomes
log

k = l o g A - \frac{E_{a}}{2.303 R T}

For the given reactions, following data is given

P \to Q k_{1} = 10^{15} e x p (\frac{- 2000}{T})

C \to D k_{2} = 10^{14} e x p (\frac{- 1000}{T})

Temperature at which

k_{1} = k_{2}

is:

Q. The rate constants

K_{1}

and

K_{2}

for two different reactions are

10^{15} e^{- \frac{1000}{T}}

and

10^{16} e^{- \frac{2000}{T}}

respectively. The temperature at which

K_{1} = K_{2}

is:

Q. The rate constants

k_{1}

and

k_{2}

for two different reactions are

10^{16} . e^{- 2000 / T}

and

10^{15} . e^{- 1000 / T}

respectively, the temperature at which

k_{1} = k_{2}

, is:

Q. For the two gaseous reactions, following data.

A \to B; k_{1} = 10^{10} e^{- 20, 000 / T}

C \to D; k_{2} = 10^{12} e^{- 24, 606 / T}

the temperature at which

k_{1}

becomes equal to

k_{2}

is:

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For the given reactions, following data is givenP→Qk1=1015exp(−2000T)C→Dk2=1014exp(−1000T)Temperature at which k1=k2 is:

The correct option is A 434.22 KAs we know,k1=1015exp(−2000T);k2=1014exp(−1000T)When k1=k2;1015exp(−2000T)=1014exp(−1000T)Or T = 434.22 K

For the given reactions, following data is given
$P \to Q k_{1} = 10^{15} e x p (\frac{- 2000}{T})$
$C \to D k_{2} = 10^{14} e x p (\frac{- 1000}{T})$
Temperature at which $k_{1} = k_{2}$ is:

The correct option is A 434.22 K
As we know,
$k_{1} = 10^{15} e x p (\frac{- 2000}{T}); k_{2} = 10^{14} e x p (\frac{- 1000}{T})$
When $k_{1} = k_{2}; 10^{15} e x p (\frac{- 2000}{T}) = 10^{14} e x p (\frac{- 1000}{T})$
Or T = 434.22 K