NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics
Q-1: Choose the correct answer. A thermodynamic state function is a quantity
(i) used to determine heat changes
(ii) whose value is independent of the path
(iii) used to determine pressure-volume work
(iv) whose value depends on temperature only
Ans:
(ii) whose value is independent of the path.
Reason:
Functions like pressure, volume and temperature depend on the state of the system only and not on the path.
Q-2: For the process to occur under adiabatic conditions, the correct condition is:
(i) ∆T = 0 (ii) ∆p = 0
(iii) q = 0 (iv) w = 0
Ans:
(iii) q = 0
Reason:
For an adiabatic process, heat transfer is zero, i.e., q = 0.
Q-3: The enthalpies of all elements in their standard states are:
(i) Unity (ii) Zero
(iii) < 0 (iv) Different for every element
Ans:
(ii) Zero
Q-4: ∆U0 of combustion of methane is – X kJ mol–1. The value of ∆H0 is
(i) =
\(\begin{array}{l}\Delta U^{\Theta }\end{array} \)
(ii) >
\(\begin{array}{l}\Delta U^{\Theta }\end{array} \)
(iii) <
\(\begin{array}{l}\Delta U^{\Theta }\end{array} \)
(iv) 0
Ans:
(iii) <
\(\begin{array}{l}\Delta U^{\Theta }\end{array} \)
Reason:
\(\begin{array}{l}\Delta H^{\Theta } = \Delta U^{\Theta } + \Delta n_{g}RT\end{array} \)
;
\(\begin{array}{l}\Delta U^{\Theta }\end{array} \)
= – Y
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
,
\(\begin{array}{l}\Delta H^{\Theta } = ( – Y) + \Delta n_{g}RT \Rightarrow \Delta H^{\Theta } < \Delta U^{\Theta }\end{array} \)
Q-5: The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are –890.3 kJ mol–1 –393.5 kJ mol–1, and –285.8 kJ mol–1, respectively. Enthalpy of the formation of CH4 (g) will be
(i) -74.8 kJ
\(\begin{array}{l}mol^{-1}\end{array} \)
(ii) -52.27 kJ
\(\begin{array}{l}mol^{-1}\end{array} \)
(iii) +74.8 kJ
\(\begin{array}{l}mol^{-1}\end{array} \)
(iv) +52 kJ
\(\begin{array}{l}mol^{-1}\end{array} \)
Ans:
(i) -74.8 kJ
\(\begin{array}{l}mol^{-1}\end{array} \)
1. CH4(g) + 2O2(g)
\(\begin{array}{l}\rightarrow\end{array} \)
CO2(g) + 2H2O(g)
\(\begin{array}{l}\Delta H = -890.3kJmol^{-1}\end{array} \)
2. C(s) + O2(g)
\(\begin{array}{l}\rightarrow\end{array} \)
CO2(g)
\(\begin{array}{l}\Delta H = -393.5kJmol^{-1}\end{array} \)
3. 2H2(g) + O2(g)
\(\begin{array}{l}\rightarrow\end{array} \)
2H2O(g)
\(\begin{array}{l}\Delta H = -285.8kJmol^{-1}\end{array} \)
C(s) + 2H2(g)
\(\begin{array}{l}\rightarrow\end{array} \)
CH4(g)
\(\begin{array}{l}\Delta _{f}H_{CH_{4}}\end{array} \)
=
\(\begin{array}{l}\Delta _{c}H_{c}\end{array} \)
+
\(\begin{array}{l}2\Delta _{f}H_{H_{2}}\end{array} \)
–
\(\begin{array}{l}\Delta _{f}H_{CO_{2}}\end{array} \)
= [ -393.5 +2(-285.8) – (-890.3)] kJ
\(\begin{array}{l}mol^{-1}\end{array} \)
= -74.8 kJ
\(\begin{array}{l}mol^{-1}\end{array} \)
Q-6: A reaction, A + B → C + D + q, is found to have a positive entropy change. The reaction will be
(i) possible at high temperature
(ii) possible only at low temperature
(iii) not possible at any temperature
(iv) possible at any temperature
Ans:
(iv) possible at any temperature
\(\begin{array}{l}\Delta G\end{array} \)
should be –ve for spontaneous reaction to occur
\(\begin{array}{l}\Delta G\end{array} \)
=
\(\begin{array}{l}\Delta H\end{array} \)
– T
\(\begin{array}{l}\Delta S\end{array} \)
As per the given question,
\(\begin{array}{l}\Delta H\end{array} \)
is –ve ( as heat is evolved)
\(\begin{array}{l}\Delta S\end{array} \)
is +ve
Therefore,
\(\begin{array}{l}\Delta G\end{array} \)
is negative
So, the reaction will be possible at any temperature.
Q-7: In a process, 701 J of heat is absorbed by a system, and 394 J of work is done by the system. What is the change in internal energy for the process?
Ans:
As per Thermodynamics 1st law,
\(\begin{array}{l}\Delta U\end{array} \)
= q + W(i);
\(\begin{array}{l}\Delta U\end{array} \)
internal energy = heat
W = work done
W = -594 J (work done by the system)
q = +701 J (+ve as heat is absorbed)
Now,
\(\begin{array}{l}\Delta U\end{array} \)
= 701 + (-594)
\(\begin{array}{l}\Delta U\end{array} \)
= 307 J
Q-8: The reaction of cyanamide, NH2CN(s), with dioxygen was carried out in a bomb calorimeter, and ∆U was found to be –742.7 kJ mol–1 at 298 K. Calculate the enthalpy change for the reaction at 298 K.
NH2CN(g) + 3/2 O2(g) → N2(g) + CO2(g) + H2O(l)
Ans:
\(\begin{array}{l}\Delta H\end{array} \)
is given by,
\(\begin{array}{l}\Delta H = \Delta U + \Delta n_{g}RT\end{array} \)
………………(1)
\(\begin{array}{l}\Delta n_{g}\end{array} \)
= change in number of moles
\(\begin{array}{l}\Delta U\end{array} \)
= change in internal energy
Here,
\(\begin{array}{l}\Delta n_{g} = \sum n_{g}(product) – \sum n_{g}(reactant)\end{array} \)
= (2 – 1.5) moles
\(\begin{array}{l}\Delta n_{g}\end{array} \)
= 0.5 moles
Here,
T =298 K
\(\begin{array}{l}\Delta U\end{array} \)
= -742.7
\(\begin{array}{l}kJmol^{-1}\end{array} \)
R =
\(\begin{array}{l}8.314\times 10^{-3}kJmol^{-1}K^{-1}\end{array} \)
Now, from (1)
\(\begin{array}{l}\Delta H = (-742.7 kJmol^{-1}) + (0.5mol)(298K)( 8.314\times 10^{-3}kJmol^{-1}K^{-1})\end{array} \)
= -742.7 + 1.2
\(\begin{array}{l}\Delta H\end{array} \)
= -741.5
\(\begin{array}{l}kJmol^{-1}\end{array} \)
Q-9: Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. The molar heat capacity of Al is 24 J mol–1 K–1.
Ans:
Expression of heat(q),
\(\begin{array}{l}q = mCP\Delta T\end{array} \)
;………………….(a)
\(\begin{array}{l}\Delta T\end{array} \)
= Change in temperature
c = molar heat capacity
m = mass of substance
From (a)
\(\begin{array}{l}q = ( \frac{60}{27}mol )(24mol^{-1}K^{-1})(20K)\end{array} \)
q = 1066.67 J = 1.067 KJ
Q-10: Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to the ice at –10.0°C. ∆fusH = 6.03 kJ mol–1 at 0°C.
\(\begin{array}{l}C_{p}[H_{2}O_{(l)}] = 75.3 J\;mol^{-1}K^{-1}\end{array} \)
\(\begin{array}{l}C_{p}[H_{2}O_{(s)}] = 36.8 J\;mol^{-1}K^{-1}\end{array} \)
Ans:
\(\begin{array}{l}\Delta H_{total}\end{array} \)
= sum of the changes given below:
(a) Energy change that occurs during the transformation of 1 mole of water from
\(\begin{array}{l}10^{\circ}C\;to\;0^{\circ}C\end{array} \)
.
(b) Energy change that occurs during the transformation of 1 mole of water at
\(\begin{array}{l}0^{\circ}C\end{array} \)
to 1 mole of ice at
\(\begin{array}{l}0^{\circ}C\end{array} \)
.
(c) Energy change that occurs during the transformation of 1 mole of ice from
\(\begin{array}{l}0^{\circ}C\;to\;(-10)^{\circ}C\end{array} \)
.
\(\begin{array}{l}\Delta H_{total} = C_{p}[H_{2}OCl]\Delta T + \Delta H_{freezing} C_{p}[H_{2}O_{l}]\Delta T\end{array} \)
= (75.3
\(\begin{array}{l}J mol^{-1}K^{-1}\end{array} \)
)(0 – 10)K + (-6.03*1000
\(\begin{array}{l}J mol^{-1}\end{array} \)
(-10-0)K
= -753
\(\begin{array}{l}J mol^{-1}\end{array} \)
– 6030
\(\begin{array}{l}J mol^{-1}\end{array} \)
– 368
\(\begin{array}{l}J mol^{-1}\end{array} \)
= -7151
\(\begin{array}{l}J mol^{-1}\end{array} \)
= -7.151
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
Thus, the required change in enthalpy for the given transformation is -7.151
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
.
Q-11 Enthalpy of combustion of carbon to CO2 is –393.5 kJ mol–1. Calculate the heat released upon the formation of 35.2 g of CO2 from carbon and dioxygen gas.
Ans:
Formation of carbon dioxide from di-oxygen and carbon gas is given as:
C(s) + O2(g) → CO2(g);
\(\begin{array}{l}\Delta _{f}H\end{array} \)
= -393.5
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
1 mole CO2 = 44g
Heat released during formation of 44g CO2 = -393.5
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
Therefore, the heat released during the formation of 35.2g of CO2 can be calculated as
=
\(\begin{array}{l}\frac{-393.5kJmol^{-1}}{44g}\times 35.2g\end{array} \)
= -314.8
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
Q-12: Enthalpies of formation of CO (g), CO2 (g), N2O (g) and N2O4 (g) are –110, – 393, 81 and 9.7 kJ mol–1, respectively. Find the value of ∆rH for the reaction:
N2O4(g) + 3CO(g) → N2O(g) + 3 CO2(g)
Ans:
“
\(\begin{array}{l}\Delta _{r}H\end{array} \)
for any reaction is defined as the difference between
\(\begin{array}{l}\Delta _{f}H\end{array} \)
value of products and
\(\begin{array}{l}\Delta _{f}H\end{array} \)
value of reactants.”
\(\begin{array}{l}\Delta _{r}H = \sum \Delta _{f}H (products) – \sum \Delta _{f}H (reactants)\end{array} \)
Now, for
N2O4(g) + 3CO(g) à N2O(g) + 3 CO2(g)
\(\begin{array}{l}\Delta _{r}H = [(\Delta _{f}H(N_{2}O) + (3\Delta _{f}H(CO_{2})) – (\Delta _{f}H(N_{2}O_{4}) + 3\Delta _{f}H(CO))]\end{array} \)
Now, substituting the given values in the above equation, we get:
\(\begin{array}{l}\Delta _{r}H\end{array} \)
= [{81
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
+ 3(-393)
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
} – {9.7
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
+ 3(-110)
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
}]
\(\begin{array}{l}\Delta _{r}H\end{array} \)
= -777.7
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
Q-13: Given N2 (g) + 3H2 (g) → 2NH3 (g) ; ∆rH0= –92.4 kJ mol–1
What is the standard enthalpy of the formation of NH3 gas?
Ans:
“Standard enthalpy of formation of a compound is the enthalpy that takes place during the formation of 1 mole of a substance in its standard form, from its constituent elements in their standard state.”
Dividing the chemical equation given in the question by 2, we get
(0.5)N2(g) + (1.5)H2(g) → 2NH3(g)
Therefore, standard enthalpy for formation of ammonia gas
= (0.5)
\(\begin{array}{l}\Delta _{r}H^{\Theta }\end{array} \)
= (0.5)(-92.4
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
)
= -46.2
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
Q-14: Calculate the standard enthalpy of formation of CH3OH(l) from the following data:
CH3OH(l) + 3/2 O2(g) → CO2(g) + 2H2O(l);
\(\begin{array}{l}\Delta _{r}H^{\Theta }\end{array} \)
= -726
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
C(g) + O2(g) → CO2(g);
\(\begin{array}{l}\Delta _{c}H_{\Theta }\end{array} \)
= -393
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
H2(g) + 1/2 O2(g) → H2O(l);
\(\begin{array}{l}\Delta _{f}H^{\Theta }\end{array} \)
= -286
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
Ans:
C(s) + 2H2O(g) + (1/2)O2(g) → CH3OH(l) …………………………(i)
CH3OH(l) can be obtained as follows,
\(\begin{array}{l}\Delta _{f}H_{\Theta }\end{array} \)
[CH3OH(l)] =
\(\begin{array}{l}\Delta _{c}H_{\Theta }\end{array} \)
2
\(\begin{array}{l}\Delta _{f}H_{\Theta }\end{array} \)
–
\(\begin{array}{l}\Delta _{r}H_{\Theta }\end{array} \)
= (-393
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
) +2(-286
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
) – (-726
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
)
= (-393 – 572 + 726)
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
= -239
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
Thus,
\(\begin{array}{l}\Delta _{f}H_{\Theta }\end{array} \)
[CH3OH(l)] = -239
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
Q-15: Calculate the enthalpy change for the process
CCl4(g) → C(g) + 4Cl(g) and determine the value of bond enthalpy for C-Cl in CCl4(g).
\(\begin{array}{l}\Delta _{vap}H^{\Theta }\end{array} \)
(CCl4) = 30.5
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
.
\(\begin{array}{l}\Delta _{f}H^{\Theta }\end{array} \)
(CCl4) = -135.5
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
.
\(\begin{array}{l}\Delta _{a}H^{\Theta }\end{array} \)
(C) = 715
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
,
\(\begin{array}{l}\Delta _{a}H^{\Theta }\end{array} \)
is a enthalpy of atomisation
\(\begin{array}{l}\Delta _{a}H^{\Theta }\end{array} \)
(Cl2) = 242
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
.
Ans:
“ The chemical equations implying the given values of enthalpies” are:
(1) CCl4(l) à CCl4(g) ;
\(\begin{array}{l}\Delta _{vap}H^{\Theta }\end{array} \)
= 30.5
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
(2) C(s) à C(g)
\(\begin{array}{l}\Delta _{a}H^{\Theta }\end{array} \)
= 715
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
(3) Cl2(g) à 2Cl(g) ;
\(\begin{array}{l}\Delta _{a}H^{\Theta }\end{array} \)
= 242
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
(4) C(g) + 4Cl(g) à CCl4(g);
\(\begin{array}{l}\Delta _{f}H^{\Theta }\end{array} \)
= -135.5
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
\(\begin{array}{l}\Delta H\end{array} \)
for the process CCl4(g) à C(g) + 4Cl(g) can be measured as:
\(\begin{array}{l}\Delta H = \Delta _{a}H^{\Theta }(C) + 2\Delta _{a}H^{\Theta }(Cl_{2}) – \Delta _{vap}H^{\Theta } – \Delta _{f}H\end{array} \)
= (715
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
) + 2(
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
) – (30.5
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
) – (-135.5
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
)
Therefore,
\(\begin{array}{l}H = 1304kJmol^{-1}\end{array} \)
The value of bond enthalpy for C-Cl in CCl4(g)
=
\(\begin{array}{l}\frac{1304}{4}kJmol^{-1}\end{array} \)
= 326
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
Q-16: For an isolated system, ∆U = 0, what will be ∆S?
Ans:
\(\begin{array}{l}\Delta U\end{array} \)
is positive ;
\(\begin{array}{l}\Delta U\end{array} \)
> 0.
As
\(\begin{array}{l}\Delta U\end{array} \)
= 0, then
\(\begin{array}{l}\Delta S\end{array} \)
will be +ve, and as a result, the reaction will be spontaneous.
Q-17: For the reaction at 298K,
2A + B → C
\(\begin{array}{l}\Delta H\end{array} \)
= 400
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
\(\begin{array}{l}\Delta H\end{array} \)
= 0.2
\(\begin{array}{l}kJ mol^{-1}K^{-1}\end{array} \)
At what temperature will the reaction become spontaneous considering
\(\begin{array}{l}\Delta S\end{array} \)
and
\(\begin{array}{l}\Delta H\end{array} \)
to be constant over the temperature range?
Ans:
Now,
\(\begin{array}{l}\Delta G = \Delta H – T\Delta S\end{array} \)
Let the given reaction is at equilibrium, then
\(\begin{array}{l}\Delta T\end{array} \)
will be:
T =
\(\begin{array}{l}(\Delta H – \Delta G)\frac{1}{\Delta S}\end{array} \)
\(\begin{array}{l}\frac{\Delta H}{\Delta S}\end{array} \)
; (
\(\begin{array}{l}\Delta G\end{array} \)
= 0 at equilibrium)
= 400
\(\begin{array}{l}kJ mol^{-1}\end{array} \)
/0.2
\(\begin{array}{l}kJ mol^{-1}K^{-1}\end{array} \)
Therefore, T = 2000K
Thus, for the spontaneous,
\(\begin{array}{l}\Delta G\end{array} \)
must be –ve and T > 2000K.
Q-18: For the reaction
2Cl(g) → Cl2(g)
What are the signs of
\(\begin{array}{l}\Delta S\end{array} \)
and
\(\begin{array}{l}\Delta H\end{array} \)
?
Ans:
\(\begin{array}{l}\Delta S\end{array} \)
and
\(\begin{array}{l}\Delta H\end{array} \)
are having negative sign.
The reaction given in the question represents the formation of Cl molecules from Cl atoms. As the formation of a bond takes place in the given reaction, energy is released. So,
\(\begin{array}{l}\Delta H\end{array} \)
is negative.
Also, 2 moles of Chlorine atoms have more randomness than 1 mole of chlorine molecule. So, the spontaneity is decreased. Thus,
\(\begin{array}{l}\Delta S\end{array} \)
is negative.
Q-19: For the reaction
2A(g) + B(g) → 2D(g)
\(\begin{array}{l}\Delta U^{\Theta }\end{array} \)
= -10.5 kJ and
\(\begin{array}{l}\Delta S^{\Theta }\end{array} \)
= -44.1
\(\begin{array}{l}JK^{-1}\end{array} \)
Calculate
\(\begin{array}{l}\Delta G^{\Theta }\end{array} \)
for the reaction, and predict whether the reaction may occur spontaneously.
Ans:
2A(g) + B(g) → 2D(g)
\(\begin{array}{l}\Delta n_{g}\end{array} \)
= 2 – 3
= -1 mole
Putting value of
\(\begin{array}{l}\Delta U^{\Theta }\end{array} \)
in expression of
\(\begin{array}{l}\Delta H\end{array} \)
:
\(\begin{array}{l}\Delta H^{\Theta } = \Delta U^{\Theta } + \Delta n_{g}RT\end{array} \)
= (-10.5KJ) – (-1)(
\(\begin{array}{l}8.314\times 10^{-3}kJK^{-1}mol^{-1}\end{array} \)
)(298K)
= -10.5kJ -2.48kJ
\(\begin{array}{l}\Delta H^{\Theta }\end{array} \)
= -12.98kJ
Putting value of
\(\begin{array}{l}\Delta S^{\Theta }\end{array} \)
and
\(\begin{array}{l}\Delta H^{\Theta }\end{array} \)
in expression of
\(\begin{array}{l}\Delta G^{\Theta }\end{array} \)
:
\(\begin{array}{l}\Delta G^{\Theta } = \Delta H^{\Theta } – T\Delta S^{\Theta }\end{array} \)
= -12.98kJ –(298K)(-44.1
\(\begin{array}{l}JK^{-1}\end{array} \)
)
= -12.98kJ +13.14kJ
\(\begin{array}{l}\Delta G^{\Theta }\end{array} \)
= 0.16kJ
As
\(\begin{array}{l}\Delta G^{\Theta }\end{array} \)
is positive, the reaction won’t occur spontaneously.
Q-20: The equilibrium constant for a reaction is 10. What will be the value of ∆G0? R = 8.314 JK–1 mol–1, T = 300 K.
Ans:
Now,
\(\begin{array}{l}\Delta G^{\Theta }\end{array} \)
=
\(\begin{array}{l}-2.303RT\ln k\end{array} \)
= (2.303)(
\(\begin{array}{l}8.314\times kJK^{-1}mol^{-1}\end{array} \)
)(300K)
\(\begin{array}{l}\log 10\end{array} \)
= -5527
\(\begin{array}{l}Jmol^{-1}\end{array} \)
= -5.527
\(\begin{array}{l}kJmol^{-1}\end{array} \)
Q-21: Comment on the thermodynamic stability of NO(g), given,
(1/2)N2(g) + (1/2)O2(g) → NO(g);
\(\begin{array}{l}\Delta _{r}H^{\Theta } = 90kJmol^{-1}\end{array} \)
NO(g) + (1/2)O2(g) → NO2(g);
\(\begin{array}{l}\Delta _{r}H^{\Theta } = -74kJmol^{-1}\end{array} \)
Ans:
The +ve value of
\(\begin{array}{l}\Delta _{r}H\end{array} \)
represents that during NO(g) formation from O2 and N2, heat is absorbed. The obtained product, NO(g), will have more energy than reactants. Thus, NO(g) is unstable.
The -ve value of
\(\begin{array}{l}\Delta _{r}H\end{array} \)
represents that during NO2(g) formation from O2(g) and NO(g), heat is evolved. The obtained product, NO2(g), gets stabilised with minimum energy.
Thus, unstable NO(g) converts into stable NO2(g).
Q-22: Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions. ∆f H0= –286 kJ mol–1.
Ans:
\(\begin{array}{l}\Delta _{r}H^{\Theta } = -286kJmol^{-1}\end{array} \)
is given, so that amount of heat is evolved during the formation of 1 mole of H2O(l).
Thus, the same heat will be absorbed by surrounding Qsurr = +286
\(\begin{array}{l}kJmol^{-1}\end{array} \)
.
Now,
\(\begin{array}{l}\Delta S_{surr}\end{array} \)
= Qsurr/7
=
\(\begin{array}{l}\frac{286kJmol^{-1}}{298K}\end{array} \)
Therefore,
\(\begin{array}{l}\Delta S_{surr} = 959.73Jmol^{-1}K^{-1}\end{array} \)
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Q1
How are NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics helpful for annual exam preparation?
NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics are created by expert faculty at ANAND CLASSES (A School Of Competitions) to help students prepare well for the annual examinations. These NCERT Solutions for Class 11 will help students to solve the problems given in the textbook comfortably. They give detailed and stepwise explanations of the problems given in the exercises of NCERT Solutions for Class 11. These solutions will help students prepare well for their upcoming annual exams by covering the entire syllabus in accordance with the NCERT syllabus.
Q2
What is thermodynamics according to NCERT Solutions for Class 11 Chemistry Chapter 6?
Thermodynamics deals with the concepts of heat and temperature and the inter-conversion of heat and other forms of energy. Thermodynamics in Physics is a branch that deals with heat, work and temperature and their relation to energy, radiation and physical properties of matter.
Q3
Are the NCERT Solutions for Class 11 Chemistry Chapter 6 at ANAND CLASSES (A School Of Competitions) provide answers to all questions present in the NCERT textbook?
The NCERT Solutions for Class 11 Chemistry Chapter 6 are useful for students as it helps them score well in the annual exams. The subject-matter experts at ANAND CLASSES (A School Of Competitions) collated detailed chapter-wise solutions for students to understand the concepts easily. These NCERT Solutions are created following the latest CBSE guidelines and are framed in accordance with the latest CBSE Syllabus; and have assembled model questions covering all the exercise questions from the textbook.
Q4.
List the Subtopics included in NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics
- Thermodynamic Terms
- The System and the Surroundings
- Types of Thermodynamic Systems
- State of the System
- Internal Energy as a State Function
- Applications
- Measurement of ΔU and ΔH: Calorimetry
- Enthalpy Change and Reaction Enthalpy
- Enthalpies for Different Types of Reactions
- Spontaneity
- Gibbs Energy Change and Equilibrium
