In a laboratory,a biochemical reaction performed outside the human body in the absence of an enzyme proceeds at a rate $10^{-6}$ times slower. If this reaction is performed in the presence of an enzyme,what will be the activation energy of the reaction?
- A$6RT$
- BExternal pressure is required
- CDifferent value from the $E_a$ of the laboratory reaction
- DCannot be determined
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The rate constant for the decomposition of $N_{2}O_{5}$ at various temperatures is given below:
$T / ^{\circ}C$ $0$ $20$ $40$ $60$ $80$ $10^{5} \times k / s^{-1}$ $0.0787$ $1.70$ $25.7$ $178$ $2140$
Draw a graph between $\ln k$ and $1 / T$ and calculate the values of $A$ and $E_{a}.$ Predict the rate constant at $30^{\circ}C$ and $50^{\circ}C$.
| $T / ^{\circ}C$ | $0$ | $20$ | $40$ | $60$ | $80$ |
| $10^{5} \times k / s^{-1}$ | $0.0787$ | $1.70$ | $25.7$ | $178$ | $2140$ |
Draw a graph between $\ln k$ and $1 / T$ and calculate the values of $A$ and $E_{a}.$ Predict the rate constant at $30^{\circ}C$ and $50^{\circ}C$.
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View SolutionFor the gaseous reactions,calculate the approximate value of temperature at which $k_1 = k_2$. $[\ln\, 10 = 2.3]$.
$A \to B$ $k_1 = 10^{15} e^{-25000 / 8.314\, T}$
$C \to D$ $k_2 = 10^{14} e^{-15000 / 8.314\, T}$
$A \to B$ $k_1 = 10^{15} e^{-25000 / 8.314\, T}$
$C \to D$ $k_2 = 10^{14} e^{-15000 / 8.314\, T}$
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View SolutionHow does the graph of the fraction of molecules versus kinetic energy change at different temperatures?
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View SolutionWrite the logarithmic form of the Arrhenius equation $k = A e^{-\frac{E_a}{RT}}$.
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View SolutionThe first order rate constant for the decomposition of ethyl iodide by the reaction $C_{2}H_{5}I_{(g)} \rightarrow C_{2}H_{4(g)} + HI_{(g)}$ at $600 \ K$ is $1.60 \times 10^{-5} \ s^{-1}$. Its energy of activation is $209 \ kJ/mol$. Calculate the rate constant of the reaction at $700 \ K$.
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