229343
At $527^{\circ} \mathrm{C}$ temperature the activation energy is 54.7 $\mathrm{KJ} / \mathrm{mole}$. The value of Arrhenius factor is $4 \times 10^{10}$. The rate constant will be
1 $12.28 \times 10^{-7}$
2 $10.76 \times 10^6$
3 $10^7$
4 $10.76 \times 10^{-6}$
Explanation:
Given that, $\mathrm{T}=527^{\circ} \mathrm{C}=527+273=800 \mathrm{~K}$ $\mathrm{E}_{\mathrm{a}}=54.7 \mathrm{~kJ} / \mathrm{mol}, \mathrm{A}=4 \times 10^{10}, \mathrm{~K}=$ ? We know that, $\begin{aligned} & K=A e^{-E_{\mathrm{a}} / R T} \\ & \log K=\log \mathrm{A}-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}} \end{aligned}$ Putting the value of these, we get - $\begin{aligned} & \log \mathrm{K}=\log \left(4 \times 10^{10}\right)-\frac{54.7 \times 10^3}{2.303 \times 8.314 \times 800} \\ & \log \mathrm{K}=10.602-3.57 \\ & \log \mathrm{K}=7.032 \\ & \mathrm{~K}=10^7 \end{aligned}$
AIIMS-25.05.2019 (Evening)
Ionic Equilibrium
229346
In an endothermic reaction $A \rightarrow B$, the activation energy is $10 \mathrm{kcal} \mathrm{mol}^{-1}$ and enthalpy of reaction is $+5 \mathrm{kcal} \mathrm{mol}^{-1}$. The activation energy for the backward reaction is
229348
In terms of Arrhenius equation, $\mathbf{k}=\mathrm{Ae}^{-\mathrm{E}_2 / \mathrm{RT}}$ the temperature dependence of rate constant (k) of a chemical reaction is written. Then, the activation energy $\left(E_a\right.$ ) of the reaction can be calculate by plotting.
1 $\log \mathrm{kvs} \frac{1}{\log \mathrm{T}}$
2 $\log \mathrm{kvs} \frac{1}{\mathrm{~T}}$
3 $\mathrm{kvs} \mathrm{T}$
4 $\mathrm{kvs} \frac{1}{\log \mathrm{T}}$
Explanation:
According to the Arrhenius equation - $\mathrm{K}=\mathrm{Ae}^{-\mathrm{E}_{\mathrm{d}} / R T}$ Taking 'In' on both side, we get - $\begin{aligned} \ln k & =\ln \mathrm{A}+\ln \mathrm{e}^{-\mathrm{E}_{\mathrm{a}} / R T} \\ \ln k & =\frac{\ln \mathrm{A}}{\mathrm{C}}-\frac{\mathrm{E}_{\mathrm{a}}}{R T} \\ \log \mathrm{k} & =\frac{\ln \mathrm{A}}{2.303}-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}} \end{aligned}$ Thus, $\log \mathrm{k}$ vs $\frac{1}{\mathrm{~T}}$
BCECE-2015
Ionic Equilibrium
229350
The Arrhenius equation gives the dependence of the rate constant of a chemical reaction of the absolute temperature and may be expressed as $K=A e^{-E_2 / R T}$. If the unit of $K$ is $s^{-1}$ the unit of A will be
1 $\mathrm{kJ} \mathrm{mol}^{-1}$
2 $\mathrm{mol}^{-1}$
3 $\mathrm{s}^{-1}$
4 $\mathrm{K}$ (kelvin)
Explanation:
According to the Arrhenius equation- $\mathrm{K}=\mathrm{Ae}^{-\mathrm{E}_2 / R T}$ Where, $A$ is the pre-exponential factor. The unit of the pre-exponential factor $\mathrm{A}$ are identical to those of the rate constant and will be vary depending on the order of the reaction. For a first order reaction. it has unit of $\mathrm{s}^{-1}$. For the reason, it is often called frequency factor.
**CG PET-2019**
Ionic Equilibrium
229352
According to Arrhenius equation. The slope of $\log K$ vs $\frac{1}{\mathrm{~T}}$ plot is.....
According to the Arrhenius equation - $\mathrm{K}=\mathrm{Ae}^{-\mathrm{E}_{\mathrm{R}} / R T}$ Taking log on both sides, we get- $\begin{aligned} \because \quad \log K & =\log \mathrm{A}-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}} \\ \log K & =\frac{-\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R} T} \frac{1}{\mathrm{~T}}+\log \mathrm{A} \end{aligned}$ When $\log K$ is plotted with $\frac{1}{\mathrm{~T}}$, the slope of will be $\frac{-E_{\mathrm{a}}}{2.303 \mathrm{R}}$ and intercept given by $\log \mathrm{A}$.
229343
At $527^{\circ} \mathrm{C}$ temperature the activation energy is 54.7 $\mathrm{KJ} / \mathrm{mole}$. The value of Arrhenius factor is $4 \times 10^{10}$. The rate constant will be
1 $12.28 \times 10^{-7}$
2 $10.76 \times 10^6$
3 $10^7$
4 $10.76 \times 10^{-6}$
Explanation:
Given that, $\mathrm{T}=527^{\circ} \mathrm{C}=527+273=800 \mathrm{~K}$ $\mathrm{E}_{\mathrm{a}}=54.7 \mathrm{~kJ} / \mathrm{mol}, \mathrm{A}=4 \times 10^{10}, \mathrm{~K}=$ ? We know that, $\begin{aligned} & K=A e^{-E_{\mathrm{a}} / R T} \\ & \log K=\log \mathrm{A}-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}} \end{aligned}$ Putting the value of these, we get - $\begin{aligned} & \log \mathrm{K}=\log \left(4 \times 10^{10}\right)-\frac{54.7 \times 10^3}{2.303 \times 8.314 \times 800} \\ & \log \mathrm{K}=10.602-3.57 \\ & \log \mathrm{K}=7.032 \\ & \mathrm{~K}=10^7 \end{aligned}$
AIIMS-25.05.2019 (Evening)
Ionic Equilibrium
229346
In an endothermic reaction $A \rightarrow B$, the activation energy is $10 \mathrm{kcal} \mathrm{mol}^{-1}$ and enthalpy of reaction is $+5 \mathrm{kcal} \mathrm{mol}^{-1}$. The activation energy for the backward reaction is
229348
In terms of Arrhenius equation, $\mathbf{k}=\mathrm{Ae}^{-\mathrm{E}_2 / \mathrm{RT}}$ the temperature dependence of rate constant (k) of a chemical reaction is written. Then, the activation energy $\left(E_a\right.$ ) of the reaction can be calculate by plotting.
1 $\log \mathrm{kvs} \frac{1}{\log \mathrm{T}}$
2 $\log \mathrm{kvs} \frac{1}{\mathrm{~T}}$
3 $\mathrm{kvs} \mathrm{T}$
4 $\mathrm{kvs} \frac{1}{\log \mathrm{T}}$
Explanation:
According to the Arrhenius equation - $\mathrm{K}=\mathrm{Ae}^{-\mathrm{E}_{\mathrm{d}} / R T}$ Taking 'In' on both side, we get - $\begin{aligned} \ln k & =\ln \mathrm{A}+\ln \mathrm{e}^{-\mathrm{E}_{\mathrm{a}} / R T} \\ \ln k & =\frac{\ln \mathrm{A}}{\mathrm{C}}-\frac{\mathrm{E}_{\mathrm{a}}}{R T} \\ \log \mathrm{k} & =\frac{\ln \mathrm{A}}{2.303}-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}} \end{aligned}$ Thus, $\log \mathrm{k}$ vs $\frac{1}{\mathrm{~T}}$
BCECE-2015
Ionic Equilibrium
229350
The Arrhenius equation gives the dependence of the rate constant of a chemical reaction of the absolute temperature and may be expressed as $K=A e^{-E_2 / R T}$. If the unit of $K$ is $s^{-1}$ the unit of A will be
1 $\mathrm{kJ} \mathrm{mol}^{-1}$
2 $\mathrm{mol}^{-1}$
3 $\mathrm{s}^{-1}$
4 $\mathrm{K}$ (kelvin)
Explanation:
According to the Arrhenius equation- $\mathrm{K}=\mathrm{Ae}^{-\mathrm{E}_2 / R T}$ Where, $A$ is the pre-exponential factor. The unit of the pre-exponential factor $\mathrm{A}$ are identical to those of the rate constant and will be vary depending on the order of the reaction. For a first order reaction. it has unit of $\mathrm{s}^{-1}$. For the reason, it is often called frequency factor.
**CG PET-2019**
Ionic Equilibrium
229352
According to Arrhenius equation. The slope of $\log K$ vs $\frac{1}{\mathrm{~T}}$ plot is.....
According to the Arrhenius equation - $\mathrm{K}=\mathrm{Ae}^{-\mathrm{E}_{\mathrm{R}} / R T}$ Taking log on both sides, we get- $\begin{aligned} \because \quad \log K & =\log \mathrm{A}-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}} \\ \log K & =\frac{-\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R} T} \frac{1}{\mathrm{~T}}+\log \mathrm{A} \end{aligned}$ When $\log K$ is plotted with $\frac{1}{\mathrm{~T}}$, the slope of will be $\frac{-E_{\mathrm{a}}}{2.303 \mathrm{R}}$ and intercept given by $\log \mathrm{A}$.
229343
At $527^{\circ} \mathrm{C}$ temperature the activation energy is 54.7 $\mathrm{KJ} / \mathrm{mole}$. The value of Arrhenius factor is $4 \times 10^{10}$. The rate constant will be
1 $12.28 \times 10^{-7}$
2 $10.76 \times 10^6$
3 $10^7$
4 $10.76 \times 10^{-6}$
Explanation:
Given that, $\mathrm{T}=527^{\circ} \mathrm{C}=527+273=800 \mathrm{~K}$ $\mathrm{E}_{\mathrm{a}}=54.7 \mathrm{~kJ} / \mathrm{mol}, \mathrm{A}=4 \times 10^{10}, \mathrm{~K}=$ ? We know that, $\begin{aligned} & K=A e^{-E_{\mathrm{a}} / R T} \\ & \log K=\log \mathrm{A}-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}} \end{aligned}$ Putting the value of these, we get - $\begin{aligned} & \log \mathrm{K}=\log \left(4 \times 10^{10}\right)-\frac{54.7 \times 10^3}{2.303 \times 8.314 \times 800} \\ & \log \mathrm{K}=10.602-3.57 \\ & \log \mathrm{K}=7.032 \\ & \mathrm{~K}=10^7 \end{aligned}$
AIIMS-25.05.2019 (Evening)
Ionic Equilibrium
229346
In an endothermic reaction $A \rightarrow B$, the activation energy is $10 \mathrm{kcal} \mathrm{mol}^{-1}$ and enthalpy of reaction is $+5 \mathrm{kcal} \mathrm{mol}^{-1}$. The activation energy for the backward reaction is
229348
In terms of Arrhenius equation, $\mathbf{k}=\mathrm{Ae}^{-\mathrm{E}_2 / \mathrm{RT}}$ the temperature dependence of rate constant (k) of a chemical reaction is written. Then, the activation energy $\left(E_a\right.$ ) of the reaction can be calculate by plotting.
1 $\log \mathrm{kvs} \frac{1}{\log \mathrm{T}}$
2 $\log \mathrm{kvs} \frac{1}{\mathrm{~T}}$
3 $\mathrm{kvs} \mathrm{T}$
4 $\mathrm{kvs} \frac{1}{\log \mathrm{T}}$
Explanation:
According to the Arrhenius equation - $\mathrm{K}=\mathrm{Ae}^{-\mathrm{E}_{\mathrm{d}} / R T}$ Taking 'In' on both side, we get - $\begin{aligned} \ln k & =\ln \mathrm{A}+\ln \mathrm{e}^{-\mathrm{E}_{\mathrm{a}} / R T} \\ \ln k & =\frac{\ln \mathrm{A}}{\mathrm{C}}-\frac{\mathrm{E}_{\mathrm{a}}}{R T} \\ \log \mathrm{k} & =\frac{\ln \mathrm{A}}{2.303}-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}} \end{aligned}$ Thus, $\log \mathrm{k}$ vs $\frac{1}{\mathrm{~T}}$
BCECE-2015
Ionic Equilibrium
229350
The Arrhenius equation gives the dependence of the rate constant of a chemical reaction of the absolute temperature and may be expressed as $K=A e^{-E_2 / R T}$. If the unit of $K$ is $s^{-1}$ the unit of A will be
1 $\mathrm{kJ} \mathrm{mol}^{-1}$
2 $\mathrm{mol}^{-1}$
3 $\mathrm{s}^{-1}$
4 $\mathrm{K}$ (kelvin)
Explanation:
According to the Arrhenius equation- $\mathrm{K}=\mathrm{Ae}^{-\mathrm{E}_2 / R T}$ Where, $A$ is the pre-exponential factor. The unit of the pre-exponential factor $\mathrm{A}$ are identical to those of the rate constant and will be vary depending on the order of the reaction. For a first order reaction. it has unit of $\mathrm{s}^{-1}$. For the reason, it is often called frequency factor.
**CG PET-2019**
Ionic Equilibrium
229352
According to Arrhenius equation. The slope of $\log K$ vs $\frac{1}{\mathrm{~T}}$ plot is.....
According to the Arrhenius equation - $\mathrm{K}=\mathrm{Ae}^{-\mathrm{E}_{\mathrm{R}} / R T}$ Taking log on both sides, we get- $\begin{aligned} \because \quad \log K & =\log \mathrm{A}-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}} \\ \log K & =\frac{-\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R} T} \frac{1}{\mathrm{~T}}+\log \mathrm{A} \end{aligned}$ When $\log K$ is plotted with $\frac{1}{\mathrm{~T}}$, the slope of will be $\frac{-E_{\mathrm{a}}}{2.303 \mathrm{R}}$ and intercept given by $\log \mathrm{A}$.
229343
At $527^{\circ} \mathrm{C}$ temperature the activation energy is 54.7 $\mathrm{KJ} / \mathrm{mole}$. The value of Arrhenius factor is $4 \times 10^{10}$. The rate constant will be
1 $12.28 \times 10^{-7}$
2 $10.76 \times 10^6$
3 $10^7$
4 $10.76 \times 10^{-6}$
Explanation:
Given that, $\mathrm{T}=527^{\circ} \mathrm{C}=527+273=800 \mathrm{~K}$ $\mathrm{E}_{\mathrm{a}}=54.7 \mathrm{~kJ} / \mathrm{mol}, \mathrm{A}=4 \times 10^{10}, \mathrm{~K}=$ ? We know that, $\begin{aligned} & K=A e^{-E_{\mathrm{a}} / R T} \\ & \log K=\log \mathrm{A}-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}} \end{aligned}$ Putting the value of these, we get - $\begin{aligned} & \log \mathrm{K}=\log \left(4 \times 10^{10}\right)-\frac{54.7 \times 10^3}{2.303 \times 8.314 \times 800} \\ & \log \mathrm{K}=10.602-3.57 \\ & \log \mathrm{K}=7.032 \\ & \mathrm{~K}=10^7 \end{aligned}$
AIIMS-25.05.2019 (Evening)
Ionic Equilibrium
229346
In an endothermic reaction $A \rightarrow B$, the activation energy is $10 \mathrm{kcal} \mathrm{mol}^{-1}$ and enthalpy of reaction is $+5 \mathrm{kcal} \mathrm{mol}^{-1}$. The activation energy for the backward reaction is
229348
In terms of Arrhenius equation, $\mathbf{k}=\mathrm{Ae}^{-\mathrm{E}_2 / \mathrm{RT}}$ the temperature dependence of rate constant (k) of a chemical reaction is written. Then, the activation energy $\left(E_a\right.$ ) of the reaction can be calculate by plotting.
1 $\log \mathrm{kvs} \frac{1}{\log \mathrm{T}}$
2 $\log \mathrm{kvs} \frac{1}{\mathrm{~T}}$
3 $\mathrm{kvs} \mathrm{T}$
4 $\mathrm{kvs} \frac{1}{\log \mathrm{T}}$
Explanation:
According to the Arrhenius equation - $\mathrm{K}=\mathrm{Ae}^{-\mathrm{E}_{\mathrm{d}} / R T}$ Taking 'In' on both side, we get - $\begin{aligned} \ln k & =\ln \mathrm{A}+\ln \mathrm{e}^{-\mathrm{E}_{\mathrm{a}} / R T} \\ \ln k & =\frac{\ln \mathrm{A}}{\mathrm{C}}-\frac{\mathrm{E}_{\mathrm{a}}}{R T} \\ \log \mathrm{k} & =\frac{\ln \mathrm{A}}{2.303}-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}} \end{aligned}$ Thus, $\log \mathrm{k}$ vs $\frac{1}{\mathrm{~T}}$
BCECE-2015
Ionic Equilibrium
229350
The Arrhenius equation gives the dependence of the rate constant of a chemical reaction of the absolute temperature and may be expressed as $K=A e^{-E_2 / R T}$. If the unit of $K$ is $s^{-1}$ the unit of A will be
1 $\mathrm{kJ} \mathrm{mol}^{-1}$
2 $\mathrm{mol}^{-1}$
3 $\mathrm{s}^{-1}$
4 $\mathrm{K}$ (kelvin)
Explanation:
According to the Arrhenius equation- $\mathrm{K}=\mathrm{Ae}^{-\mathrm{E}_2 / R T}$ Where, $A$ is the pre-exponential factor. The unit of the pre-exponential factor $\mathrm{A}$ are identical to those of the rate constant and will be vary depending on the order of the reaction. For a first order reaction. it has unit of $\mathrm{s}^{-1}$. For the reason, it is often called frequency factor.
**CG PET-2019**
Ionic Equilibrium
229352
According to Arrhenius equation. The slope of $\log K$ vs $\frac{1}{\mathrm{~T}}$ plot is.....
According to the Arrhenius equation - $\mathrm{K}=\mathrm{Ae}^{-\mathrm{E}_{\mathrm{R}} / R T}$ Taking log on both sides, we get- $\begin{aligned} \because \quad \log K & =\log \mathrm{A}-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}} \\ \log K & =\frac{-\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R} T} \frac{1}{\mathrm{~T}}+\log \mathrm{A} \end{aligned}$ When $\log K$ is plotted with $\frac{1}{\mathrm{~T}}$, the slope of will be $\frac{-E_{\mathrm{a}}}{2.303 \mathrm{R}}$ and intercept given by $\log \mathrm{A}$.
229343
At $527^{\circ} \mathrm{C}$ temperature the activation energy is 54.7 $\mathrm{KJ} / \mathrm{mole}$. The value of Arrhenius factor is $4 \times 10^{10}$. The rate constant will be
1 $12.28 \times 10^{-7}$
2 $10.76 \times 10^6$
3 $10^7$
4 $10.76 \times 10^{-6}$
Explanation:
Given that, $\mathrm{T}=527^{\circ} \mathrm{C}=527+273=800 \mathrm{~K}$ $\mathrm{E}_{\mathrm{a}}=54.7 \mathrm{~kJ} / \mathrm{mol}, \mathrm{A}=4 \times 10^{10}, \mathrm{~K}=$ ? We know that, $\begin{aligned} & K=A e^{-E_{\mathrm{a}} / R T} \\ & \log K=\log \mathrm{A}-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}} \end{aligned}$ Putting the value of these, we get - $\begin{aligned} & \log \mathrm{K}=\log \left(4 \times 10^{10}\right)-\frac{54.7 \times 10^3}{2.303 \times 8.314 \times 800} \\ & \log \mathrm{K}=10.602-3.57 \\ & \log \mathrm{K}=7.032 \\ & \mathrm{~K}=10^7 \end{aligned}$
AIIMS-25.05.2019 (Evening)
Ionic Equilibrium
229346
In an endothermic reaction $A \rightarrow B$, the activation energy is $10 \mathrm{kcal} \mathrm{mol}^{-1}$ and enthalpy of reaction is $+5 \mathrm{kcal} \mathrm{mol}^{-1}$. The activation energy for the backward reaction is
229348
In terms of Arrhenius equation, $\mathbf{k}=\mathrm{Ae}^{-\mathrm{E}_2 / \mathrm{RT}}$ the temperature dependence of rate constant (k) of a chemical reaction is written. Then, the activation energy $\left(E_a\right.$ ) of the reaction can be calculate by plotting.
1 $\log \mathrm{kvs} \frac{1}{\log \mathrm{T}}$
2 $\log \mathrm{kvs} \frac{1}{\mathrm{~T}}$
3 $\mathrm{kvs} \mathrm{T}$
4 $\mathrm{kvs} \frac{1}{\log \mathrm{T}}$
Explanation:
According to the Arrhenius equation - $\mathrm{K}=\mathrm{Ae}^{-\mathrm{E}_{\mathrm{d}} / R T}$ Taking 'In' on both side, we get - $\begin{aligned} \ln k & =\ln \mathrm{A}+\ln \mathrm{e}^{-\mathrm{E}_{\mathrm{a}} / R T} \\ \ln k & =\frac{\ln \mathrm{A}}{\mathrm{C}}-\frac{\mathrm{E}_{\mathrm{a}}}{R T} \\ \log \mathrm{k} & =\frac{\ln \mathrm{A}}{2.303}-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}} \end{aligned}$ Thus, $\log \mathrm{k}$ vs $\frac{1}{\mathrm{~T}}$
BCECE-2015
Ionic Equilibrium
229350
The Arrhenius equation gives the dependence of the rate constant of a chemical reaction of the absolute temperature and may be expressed as $K=A e^{-E_2 / R T}$. If the unit of $K$ is $s^{-1}$ the unit of A will be
1 $\mathrm{kJ} \mathrm{mol}^{-1}$
2 $\mathrm{mol}^{-1}$
3 $\mathrm{s}^{-1}$
4 $\mathrm{K}$ (kelvin)
Explanation:
According to the Arrhenius equation- $\mathrm{K}=\mathrm{Ae}^{-\mathrm{E}_2 / R T}$ Where, $A$ is the pre-exponential factor. The unit of the pre-exponential factor $\mathrm{A}$ are identical to those of the rate constant and will be vary depending on the order of the reaction. For a first order reaction. it has unit of $\mathrm{s}^{-1}$. For the reason, it is often called frequency factor.
**CG PET-2019**
Ionic Equilibrium
229352
According to Arrhenius equation. The slope of $\log K$ vs $\frac{1}{\mathrm{~T}}$ plot is.....
According to the Arrhenius equation - $\mathrm{K}=\mathrm{Ae}^{-\mathrm{E}_{\mathrm{R}} / R T}$ Taking log on both sides, we get- $\begin{aligned} \because \quad \log K & =\log \mathrm{A}-\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}} \\ \log K & =\frac{-\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R} T} \frac{1}{\mathrm{~T}}+\log \mathrm{A} \end{aligned}$ When $\log K$ is plotted with $\frac{1}{\mathrm{~T}}$, the slope of will be $\frac{-E_{\mathrm{a}}}{2.303 \mathrm{R}}$ and intercept given by $\log \mathrm{A}$.
