229066
One mole of ethanol is produced reacting graphite, $\mathrm{H}_{2}$ and $\mathrm{O}_{2}$ together. The standard enthalpy of formation is $-277.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Calculate the standard enthalpy of the reaction when 4 moles of graphite is involved
1 -277.7
2 -555.4
3 -138.85
4 -69.42
5 -1110.8
Explanation:
When graphite reacted with $\mathrm{H}_{2}$ and $\mathrm{O}_{2}$ then one mole of ethanol is formed. The chemical reaction is given below- $2 \mathrm{C}(\mathrm{s})+3 \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)$ $\Delta \mathrm{H}_{\mathrm{f}}=-277.7 \mathrm{~kJ} / \mathrm{mol}$ If 2 moles of graphite is involved, the standard enthalpy of reaction is $-277.7 \mathrm{~kJ} / \mathrm{mol}$. When 4 moles of graphite is involved then standard enthalpy of reaction will be $2 \times(-277.7 \mathrm{~kJ} / \mathrm{mol})=-555.4 \mathrm{~kJ} / \mathrm{mol}$
Kerala-CEE-2018
Chemical Equilibrium
229068
From the Ellingham graphs on carbon, which of the following statements is false?
1 $\mathrm{CO}$ reduces $\mathrm{Fe}_{2} \mathrm{O}_{3}$ to $\mathrm{Fe}$ at less than $983 \mathrm{~K}$
2 $\mathrm{CO}$ is less stable than $\mathrm{CO}_{2}$ at more than 983 $\mathrm{K}$
3 $\mathrm{CO}$ reduces $\mathrm{Fe}_{2} \mathrm{O}_{3}$ to $\mathrm{Fe}$ in the reduction zone of blast furnace
4 $\mathrm{CO}_{2}$ is more stable than $\mathrm{CO}$ at less then 983 $\mathrm{K}$
Explanation:
$\mathrm{CO}_{2}$ is less stable than $\mathrm{CO}$ at more than $983 \mathrm{~K}$. At these temperature, the $\Delta \mathrm{G}$ value of formation of CO is more negative than the $\Delta \mathrm{G}$ value of formation of $\mathrm{CO}_{2}$. More negative, the value of $\Delta \mathrm{G}$, the greater is the tendency of the element to combine with oxygen.
Karnataka-CET-2012
Chemical Equilibrium
229074
A chemical reaction is catalysed by ' $\mathrm{X}$ ' therefore, ' $\mathrm{X}$ '
1 increases the activation energy
2 does not affect the equilibrium constant of the reaction
3 decreases the enthalpy of the reaction
4 decreases the velocity constant of the reaction
Explanation:
A catalyst increases the rate of the chemical reaction by making available a new low energy pathway for the conversion of reactants to products. It increases the rate of forward and reverse reactions that pass through the same transition state and does not affect equilibrium. Catalyst lowers the activation energy for the forward and reverse reactions by exactly the same amount.
CG PET -2006
Chemical Equilibrium
229073
The reactions with low activation energy are always
1 adiabatic
2 slow
3 non-spontaneous
4 fast
Explanation:
The reactions with low activation energies are always fast where as the reactions with high activation energy are always slow.
229066
One mole of ethanol is produced reacting graphite, $\mathrm{H}_{2}$ and $\mathrm{O}_{2}$ together. The standard enthalpy of formation is $-277.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Calculate the standard enthalpy of the reaction when 4 moles of graphite is involved
1 -277.7
2 -555.4
3 -138.85
4 -69.42
5 -1110.8
Explanation:
When graphite reacted with $\mathrm{H}_{2}$ and $\mathrm{O}_{2}$ then one mole of ethanol is formed. The chemical reaction is given below- $2 \mathrm{C}(\mathrm{s})+3 \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)$ $\Delta \mathrm{H}_{\mathrm{f}}=-277.7 \mathrm{~kJ} / \mathrm{mol}$ If 2 moles of graphite is involved, the standard enthalpy of reaction is $-277.7 \mathrm{~kJ} / \mathrm{mol}$. When 4 moles of graphite is involved then standard enthalpy of reaction will be $2 \times(-277.7 \mathrm{~kJ} / \mathrm{mol})=-555.4 \mathrm{~kJ} / \mathrm{mol}$
Kerala-CEE-2018
Chemical Equilibrium
229068
From the Ellingham graphs on carbon, which of the following statements is false?
1 $\mathrm{CO}$ reduces $\mathrm{Fe}_{2} \mathrm{O}_{3}$ to $\mathrm{Fe}$ at less than $983 \mathrm{~K}$
2 $\mathrm{CO}$ is less stable than $\mathrm{CO}_{2}$ at more than 983 $\mathrm{K}$
3 $\mathrm{CO}$ reduces $\mathrm{Fe}_{2} \mathrm{O}_{3}$ to $\mathrm{Fe}$ in the reduction zone of blast furnace
4 $\mathrm{CO}_{2}$ is more stable than $\mathrm{CO}$ at less then 983 $\mathrm{K}$
Explanation:
$\mathrm{CO}_{2}$ is less stable than $\mathrm{CO}$ at more than $983 \mathrm{~K}$. At these temperature, the $\Delta \mathrm{G}$ value of formation of CO is more negative than the $\Delta \mathrm{G}$ value of formation of $\mathrm{CO}_{2}$. More negative, the value of $\Delta \mathrm{G}$, the greater is the tendency of the element to combine with oxygen.
Karnataka-CET-2012
Chemical Equilibrium
229074
A chemical reaction is catalysed by ' $\mathrm{X}$ ' therefore, ' $\mathrm{X}$ '
1 increases the activation energy
2 does not affect the equilibrium constant of the reaction
3 decreases the enthalpy of the reaction
4 decreases the velocity constant of the reaction
Explanation:
A catalyst increases the rate of the chemical reaction by making available a new low energy pathway for the conversion of reactants to products. It increases the rate of forward and reverse reactions that pass through the same transition state and does not affect equilibrium. Catalyst lowers the activation energy for the forward and reverse reactions by exactly the same amount.
CG PET -2006
Chemical Equilibrium
229073
The reactions with low activation energy are always
1 adiabatic
2 slow
3 non-spontaneous
4 fast
Explanation:
The reactions with low activation energies are always fast where as the reactions with high activation energy are always slow.
229066
One mole of ethanol is produced reacting graphite, $\mathrm{H}_{2}$ and $\mathrm{O}_{2}$ together. The standard enthalpy of formation is $-277.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Calculate the standard enthalpy of the reaction when 4 moles of graphite is involved
1 -277.7
2 -555.4
3 -138.85
4 -69.42
5 -1110.8
Explanation:
When graphite reacted with $\mathrm{H}_{2}$ and $\mathrm{O}_{2}$ then one mole of ethanol is formed. The chemical reaction is given below- $2 \mathrm{C}(\mathrm{s})+3 \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)$ $\Delta \mathrm{H}_{\mathrm{f}}=-277.7 \mathrm{~kJ} / \mathrm{mol}$ If 2 moles of graphite is involved, the standard enthalpy of reaction is $-277.7 \mathrm{~kJ} / \mathrm{mol}$. When 4 moles of graphite is involved then standard enthalpy of reaction will be $2 \times(-277.7 \mathrm{~kJ} / \mathrm{mol})=-555.4 \mathrm{~kJ} / \mathrm{mol}$
Kerala-CEE-2018
Chemical Equilibrium
229068
From the Ellingham graphs on carbon, which of the following statements is false?
1 $\mathrm{CO}$ reduces $\mathrm{Fe}_{2} \mathrm{O}_{3}$ to $\mathrm{Fe}$ at less than $983 \mathrm{~K}$
2 $\mathrm{CO}$ is less stable than $\mathrm{CO}_{2}$ at more than 983 $\mathrm{K}$
3 $\mathrm{CO}$ reduces $\mathrm{Fe}_{2} \mathrm{O}_{3}$ to $\mathrm{Fe}$ in the reduction zone of blast furnace
4 $\mathrm{CO}_{2}$ is more stable than $\mathrm{CO}$ at less then 983 $\mathrm{K}$
Explanation:
$\mathrm{CO}_{2}$ is less stable than $\mathrm{CO}$ at more than $983 \mathrm{~K}$. At these temperature, the $\Delta \mathrm{G}$ value of formation of CO is more negative than the $\Delta \mathrm{G}$ value of formation of $\mathrm{CO}_{2}$. More negative, the value of $\Delta \mathrm{G}$, the greater is the tendency of the element to combine with oxygen.
Karnataka-CET-2012
Chemical Equilibrium
229074
A chemical reaction is catalysed by ' $\mathrm{X}$ ' therefore, ' $\mathrm{X}$ '
1 increases the activation energy
2 does not affect the equilibrium constant of the reaction
3 decreases the enthalpy of the reaction
4 decreases the velocity constant of the reaction
Explanation:
A catalyst increases the rate of the chemical reaction by making available a new low energy pathway for the conversion of reactants to products. It increases the rate of forward and reverse reactions that pass through the same transition state and does not affect equilibrium. Catalyst lowers the activation energy for the forward and reverse reactions by exactly the same amount.
CG PET -2006
Chemical Equilibrium
229073
The reactions with low activation energy are always
1 adiabatic
2 slow
3 non-spontaneous
4 fast
Explanation:
The reactions with low activation energies are always fast where as the reactions with high activation energy are always slow.
NEET Test Series from KOTA - 10 Papers In MS WORD
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Chemical Equilibrium
229066
One mole of ethanol is produced reacting graphite, $\mathrm{H}_{2}$ and $\mathrm{O}_{2}$ together. The standard enthalpy of formation is $-277.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Calculate the standard enthalpy of the reaction when 4 moles of graphite is involved
1 -277.7
2 -555.4
3 -138.85
4 -69.42
5 -1110.8
Explanation:
When graphite reacted with $\mathrm{H}_{2}$ and $\mathrm{O}_{2}$ then one mole of ethanol is formed. The chemical reaction is given below- $2 \mathrm{C}(\mathrm{s})+3 \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)$ $\Delta \mathrm{H}_{\mathrm{f}}=-277.7 \mathrm{~kJ} / \mathrm{mol}$ If 2 moles of graphite is involved, the standard enthalpy of reaction is $-277.7 \mathrm{~kJ} / \mathrm{mol}$. When 4 moles of graphite is involved then standard enthalpy of reaction will be $2 \times(-277.7 \mathrm{~kJ} / \mathrm{mol})=-555.4 \mathrm{~kJ} / \mathrm{mol}$
Kerala-CEE-2018
Chemical Equilibrium
229068
From the Ellingham graphs on carbon, which of the following statements is false?
1 $\mathrm{CO}$ reduces $\mathrm{Fe}_{2} \mathrm{O}_{3}$ to $\mathrm{Fe}$ at less than $983 \mathrm{~K}$
2 $\mathrm{CO}$ is less stable than $\mathrm{CO}_{2}$ at more than 983 $\mathrm{K}$
3 $\mathrm{CO}$ reduces $\mathrm{Fe}_{2} \mathrm{O}_{3}$ to $\mathrm{Fe}$ in the reduction zone of blast furnace
4 $\mathrm{CO}_{2}$ is more stable than $\mathrm{CO}$ at less then 983 $\mathrm{K}$
Explanation:
$\mathrm{CO}_{2}$ is less stable than $\mathrm{CO}$ at more than $983 \mathrm{~K}$. At these temperature, the $\Delta \mathrm{G}$ value of formation of CO is more negative than the $\Delta \mathrm{G}$ value of formation of $\mathrm{CO}_{2}$. More negative, the value of $\Delta \mathrm{G}$, the greater is the tendency of the element to combine with oxygen.
Karnataka-CET-2012
Chemical Equilibrium
229074
A chemical reaction is catalysed by ' $\mathrm{X}$ ' therefore, ' $\mathrm{X}$ '
1 increases the activation energy
2 does not affect the equilibrium constant of the reaction
3 decreases the enthalpy of the reaction
4 decreases the velocity constant of the reaction
Explanation:
A catalyst increases the rate of the chemical reaction by making available a new low energy pathway for the conversion of reactants to products. It increases the rate of forward and reverse reactions that pass through the same transition state and does not affect equilibrium. Catalyst lowers the activation energy for the forward and reverse reactions by exactly the same amount.
CG PET -2006
Chemical Equilibrium
229073
The reactions with low activation energy are always
1 adiabatic
2 slow
3 non-spontaneous
4 fast
Explanation:
The reactions with low activation energies are always fast where as the reactions with high activation energy are always slow.
