ChemistryQ101–106 of 206 questions
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101
ChemistryMediumMCQTS EAMCET · 2019
Find the heat required to convert $10 \ g$ of ice at $0.0^{\circ} C$ into water at $30^{\circ} C$. (Enthalpy of fusion of ice $= 333.5 \ J \ g^{-1}$,$C_p$ of water $= 4.18 \ J \ g^{-1} \ K^{-1}$) (in $kJ$)
A
$4.0$
B
$5.0$
C
$3.59$
D
$4.59$
Solution
(D) The process involves two steps: melting the ice at $0^{\circ} C$ and heating the resulting water from $0^{\circ} C$ to $30^{\circ} C$.
Given:
Mass $(m) = 10 \ g$
Enthalpy of fusion $(L_f) = 333.5 \ J \ g^{-1}$
Specific heat capacity of water $(C_p) = 4.18 \ J \ g^{-1} \ K^{-1}$
Change in temperature $(\Delta T) = 30^{\circ} C = 30 \ K$
Total heat required $(Q) = Q_{fusion} + Q_{heating}$
$Q = (m \times L_f) + (m \times C_p \times \Delta T)$
$Q = (10 \ g \times 333.5 \ J \ g^{-1}) + (10 \ g \times 4.18 \ J \ g^{-1} \ K^{-1} \times 30 \ K)$
$Q = 3335 \ J + 1254 \ J$
$Q = 4589 \ J = 4.589 \ kJ \approx 4.59 \ kJ$
Thus,the correct option is $D$.
Given:
Mass $(m) = 10 \ g$
Enthalpy of fusion $(L_f) = 333.5 \ J \ g^{-1}$
Specific heat capacity of water $(C_p) = 4.18 \ J \ g^{-1} \ K^{-1}$
Change in temperature $(\Delta T) = 30^{\circ} C = 30 \ K$
Total heat required $(Q) = Q_{fusion} + Q_{heating}$
$Q = (m \times L_f) + (m \times C_p \times \Delta T)$
$Q = (10 \ g \times 333.5 \ J \ g^{-1}) + (10 \ g \times 4.18 \ J \ g^{-1} \ K^{-1} \times 30 \ K)$
$Q = 3335 \ J + 1254 \ J$
$Q = 4589 \ J = 4.589 \ kJ \approx 4.59 \ kJ$
Thus,the correct option is $D$.
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102
ChemistryMediumMCQTS EAMCET · 2019
The heat of combustion $\left(kJ \ mol^{-1}\right)$ is highest for
A
$H_{2(g)}$
B
$H_2O_{(l)}$
C
$LPG$
D
$CH_{4(g)}$
Solution
(C) The heat change when one mole of a substance is completely burnt in oxygen is called the enthalpy of combustion.
$LPG$ (Liquefied Petroleum Gas) is a mixture of hydrocarbons like propane $(C_3H_8)$ and butane $(C_4H_{10})$.
As the number of carbon atoms in the hydrocarbon chain increases,the amount of energy released upon complete combustion increases significantly.
Since $H_2O$ is already a product of combustion,it has no heat of combustion.
Comparing $H_2$,$CH_4$,and $LPG$,$LPG$ has the highest molecular weight and carbon content,resulting in the highest heat of combustion per mole.
Thus,option $(C)$ is correct.
$LPG$ (Liquefied Petroleum Gas) is a mixture of hydrocarbons like propane $(C_3H_8)$ and butane $(C_4H_{10})$.
As the number of carbon atoms in the hydrocarbon chain increases,the amount of energy released upon complete combustion increases significantly.
Since $H_2O$ is already a product of combustion,it has no heat of combustion.
Comparing $H_2$,$CH_4$,and $LPG$,$LPG$ has the highest molecular weight and carbon content,resulting in the highest heat of combustion per mole.
Thus,option $(C)$ is correct.
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103
ChemistryMediumMCQTS EAMCET · 2019
Enthalpy change for freezing of $1 \ g$ of water at $1 \ bar$ and $0^{\circ} C$ is $334 \ J$. Calculate the internal energy change in $J$ when $1 \ g$ of water is converted into ice?
A
$205$
B
$334$
C
$0$
D
$668$
Solution
(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + P\Delta V$.
At $0^{\circ} C$ and $1 \ bar$,the change in volume $(\Delta V)$ for the freezing of $1 \ g$ of water is extremely small and negligible.
Therefore,$\Delta H \approx \Delta U$.
Given that the enthalpy change for freezing $1 \ g$ of water is $334 \ J$,the internal energy change is also approximately $334 \ J$.
At $0^{\circ} C$ and $1 \ bar$,the change in volume $(\Delta V)$ for the freezing of $1 \ g$ of water is extremely small and negligible.
Therefore,$\Delta H \approx \Delta U$.
Given that the enthalpy change for freezing $1 \ g$ of water is $334 \ J$,the internal energy change is also approximately $334 \ J$.
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104
ChemistryMediumMCQTS EAMCET · 2019
What is the standard enthalpy of reaction (in $kJ$) when two moles of $Fe_2O_{3(s)}$ react with $H_2$ gas to give $Fe$ metal? $\Delta H_f^{\circ}$ of $Fe_2O_{3(s)}$ and $H_2O_{(l)}$ are $-824.2$ and $-285.83 \ kJ \ mol^{-1}$ respectively.
A
$-66.58$
B
$-33.3$
C
$-538.37$
D
$-1110.03$
Solution
(A) The balanced chemical equation for the reaction of $Fe_2O_3$ with $H_2$ is: $Fe_2O_3(s) + 3H_2(g) \rightarrow 2Fe(s) + 3H_2O(l)$.
For $2$ moles of $Fe_2O_3$,the equation is: $2Fe_2O_3(s) + 6H_2(g) \rightarrow 4Fe(s) + 6H_2O(l)$.
The standard enthalpy of reaction is given by: $\Delta H_R^{\circ} = \sum \Delta H_f^{\circ}(\text{Products}) - \sum \Delta H_f^{\circ}(\text{Reactants})$.
Given $\Delta H_f^{\circ}(Fe_2O_3) = -824.2 \ kJ \ mol^{-1}$,$\Delta H_f^{\circ}(H_2O) = -285.83 \ kJ \ mol^{-1}$,and $\Delta H_f^{\circ}(Fe) = \Delta H_f^{\circ}(H_2) = 0 \ kJ \ mol^{-1}$.
Substituting the values: $\Delta H_R^{\circ} = [4(0) + 6(-285.83)] - [2(-824.2) + 6(0)]$.
$\Delta H_R^{\circ} = -1714.98 - (-1648.4) = -66.58 \ kJ$.
Thus,the correct option is $(A)$.
For $2$ moles of $Fe_2O_3$,the equation is: $2Fe_2O_3(s) + 6H_2(g) \rightarrow 4Fe(s) + 6H_2O(l)$.
The standard enthalpy of reaction is given by: $\Delta H_R^{\circ} = \sum \Delta H_f^{\circ}(\text{Products}) - \sum \Delta H_f^{\circ}(\text{Reactants})$.
Given $\Delta H_f^{\circ}(Fe_2O_3) = -824.2 \ kJ \ mol^{-1}$,$\Delta H_f^{\circ}(H_2O) = -285.83 \ kJ \ mol^{-1}$,and $\Delta H_f^{\circ}(Fe) = \Delta H_f^{\circ}(H_2) = 0 \ kJ \ mol^{-1}$.
Substituting the values: $\Delta H_R^{\circ} = [4(0) + 6(-285.83)] - [2(-824.2) + 6(0)]$.
$\Delta H_R^{\circ} = -1714.98 - (-1648.4) = -66.58 \ kJ$.
Thus,the correct option is $(A)$.
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105
ChemistryEasyMCQTS EAMCET · 2019
The increase in entropy in $J K^{-1}$ of a substance when it absorbs $1 \ kJ$ of heat energy at $3 \ K$ is
A
$3.33$
B
$333.3$
C
$0.333$
D
$0.0333$
Solution
(B) Given,
$Heat \ (q) = 1 \ kJ = 1000 \ J$
$Temperature \ (T) = 3 \ K$
The change in entropy is given by the formula:
$\Delta S = \frac{q}{T}$
Substituting the values:
$\Delta S = \frac{1000 \ J}{3 \ K} = 333.3 \ J K^{-1}$
$Heat \ (q) = 1 \ kJ = 1000 \ J$
$Temperature \ (T) = 3 \ K$
The change in entropy is given by the formula:
$\Delta S = \frac{q}{T}$
Substituting the values:
$\Delta S = \frac{1000 \ J}{3 \ K} = 333.3 \ J K^{-1}$
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106
ChemistryMediumMCQTS EAMCET · 2019
For a chemical reaction,the standard Gibbs energy change,$\Delta G^{\circ}$ is $-7.64 \times 10^4 \ J \ mol^{-1}$. What is the value of equilibrium constant $(K)$?
A
$K=1$
B
$K > 1$
C
$K < 1$
D
$K=0$
Solution
(B) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the equation: $\Delta G^{\circ} = -RT \ln K$ or $\Delta G^{\circ} = -2.303 RT \log K$.
Given that $\Delta G^{\circ} = -7.64 \times 10^4 \ J \ mol^{-1}$.
Since $\Delta G^{\circ}$ is negative,the reaction is spontaneous in the forward direction.
For a spontaneous reaction,the equilibrium constant $K$ must be greater than $1$.
Mathematically,$\log K = -\frac{\Delta G^{\circ}}{2.303 RT} = -\frac{-7.64 \times 10^4}{2.303 \times 8.314 \times 298} > 0$.
Since $\log K > 0$,it implies $K > 10^0$,so $K > 1$.
Thus,option $(B)$ is correct.
Given that $\Delta G^{\circ} = -7.64 \times 10^4 \ J \ mol^{-1}$.
Since $\Delta G^{\circ}$ is negative,the reaction is spontaneous in the forward direction.
For a spontaneous reaction,the equilibrium constant $K$ must be greater than $1$.
Mathematically,$\log K = -\frac{\Delta G^{\circ}}{2.303 RT} = -\frac{-7.64 \times 10^4}{2.303 \times 8.314 \times 298} > 0$.
Since $\log K > 0$,it implies $K > 10^0$,so $K > 1$.
Thus,option $(B)$ is correct.
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