For the two gaseous reactions- following data is given -A - B - k 1-1010 e -20000 - TC - D - k 2-1012 e -24606 - T the temperature at which k 1 becomes equal to k 2 is-A- 400 KB- 1000 K- 800 KD- 1500 K

The correct option is B 1000 KA→ B ; k_1 = 10^10 e^ 20000 /T C → D ; k_2= 10^12 e^ 24606/T when k_1 = k_2 10^10 e^ 20000/T = 10^12e^ 24606 /T e^4606/T= 100 46 ...

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Byju's Answer

Standard XII

Chemistry

Internal Energy

For the two g...

Question

For the two gaseous reactions, following data is given :
$A \to B; k_{1} = 10^{10} e^{- 20000 / T}$
$C \to D; k_{2} = 10^{12} e^{- 24606 / T}$ the temperature at which $k_{1}$ becomes equal to $k_{2}$ is:

400 K

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1000 K

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800 K

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1500 K

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Solution

The correct option is B $1000 K$
$A \to B; k_{1} = 10^{10} e^{- 20000 / T}$ $C \to D; k_{2} = 10^{12} e^{- 24606 / T}$
when $k_{1} = k_{2}$
$10^{10} e^{- 20000 / T} = 10^{12} e^{- 24606 / T}$
$e^{4606 / T} = 100$
$\frac{4606}{T} = 2.303 log 100 = 2.303 \times 2$
$∴ T = \frac{4606}{2.303 \times 2} = 1000 K$

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For the two gaseous reactions, following data is given : A→B;k1=1010e−20000/T C→D;k2=1012e−24606/T the temperature at which k1 ​becomes equal to k2 ​ is:

The correct option is B 1000 KA→B;k1=1010e−20000/T C→D;k2=1012e−24606/T when k1=k2 1010e−20000/T=1012e−24606/T e4606/T=100 4606T=2.303 log 100=2.303×2 ∴ T=46062.303×2=1000 K

For the two gaseous reactions, following data is given :
$A \to B; k_{1} = 10^{10} e^{- 20000 / T}$
$C \to D; k_{2} = 10^{12} e^{- 24606 / T}$ the temperature at which $k_{1}$ becomes equal to $k_{2}$ is: