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Practice each question, then open the answer or worked solution.
Which quantity out of ΔrG and ΔrG° will be zero at equilibrium?
Ans: Gibbs energy for a reaction in which all reactants and products are in standard state. ΔrG° is related to the equilibrium constant of the reaction as follows
ΔrG = ArG° + RT In K
At equilibrium, 0 = ΔrG° + RT In A– ({ΔrG = 0) or ΔrG° =-RT lnK
ΔrG° = 0 when K= 1
For all other values of K, ArG° will be non-zero.
Predict the change in internal energy for an isolated system at constant volume.
Ans: For an isolated system w = 0, q = 0
Since ΔU= q + w = 0 + 0 = 0, ΔU= 0
Although heat is a path function, heat absorbed by the system under certain specific conditions is independent of path. What are those conditions? Explain.
Ans : At constant volume
q = ΔU + (-w)
-w = pΔ q = AU + pΔV
ΔV = 0 (at constant volume)
Hence, qv = ΔU + 0 = ΔU= change in internal energy at constant pressure, qp = AU + pΔV
Since ΔU + pΔV=ΔH
=> qp = ΔH change in enthalpy
Hence, at constant volume and at constant pressure, heat change is a state function because it is equal to ΔU and ΔH respectively which are state functions.
Expansion of a gas in vacuum is called free expansion. Calculate the work done and the change in internal energy when 1 litre of ideal gas expands isothermally into vacuum until its total volume is 5 litre.
Ans: During free expansion, external pressure is zero, so Work done, w = -pextΔV
= -0(5 – 1) = 0
Since the gas is expanding isothermally, therefore, q = 0
ΔU = q + w =0+0=0
Heat capacity (CP) is an extensive property but specific heat (c) is an intensive property. What will be the relation between Cp and c for 1 mol of water?
Ans: For water, molar heat capacity = 18 x Specific heat or Cp = 18 x c
But, specific heat,
C = 4.18 J g-1 K-1 Heat capacity,
Cp = 18 x 4.18 JK 1 = 75.24 JK-1
The difference between Cp and Cv can be derived using the empirical relation H = U + pV. Calculate the difference between Cp and Cv for 10 moles of an ideal gas.
Ans: Given that, Cv = heat capacity at constant volume,
Cp = heat capacity at constant pressure
Difference between Cp and Cv is equal to gas constant (R).
.’. Cp – Cv = nR (where, n = no. of moles)
= 10 x 8.314 = 83.14J
If the combustion of 1 g of graphite produces 20.7 kJ of heat, what will be the molar enthalpy change? Give the significance of the sign also.
Ans: Molar enthalpy change for graphite (ΔH)
= enthalpy change for 1 g x molar mass of C = -20.7×12 = -2.48 x 102 kJ mol-1
Since the sign of ΔH = -ve, it is an exothermic reaction.
The net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus the amount of energy required to form all the bonds in the product molecules. What will be the enthalpy change for the following reaction?
H2(g) + Br2(g)→2HBr(g)
Given that the bond energy of H2, Br2 and HBr is 4.35 kJ mol-1,192 kJ mol-1 and 368 kJ mol -1 respectively.
Ans: $\Delta $rH$\circ $=$\Sigma $B.E(reactant)-$\Sigma $B.E(product)
=B.E.H2 + BEBR2 -2 $\times $B.E>HBr
= 435+192-2($\times $368)
= 109KJmol-1
The enthalpy of vaporization of CCl4 is 30.5 kJ mol–1 . Calculate the heat required for the vaporization of 284 g of CCl4 at constant pressure. (Molar mass of CCl4 = 154 g mol–1).
Ans: As per the information provided in the question, for one mole of CCl4(154 g), the heat of vaporisation required is 30.5 kJ/mol .
Hence for the vaporisation of 284 g of CCl4, we require:
$\frac{~~~284~~~}{154}\times $30.5
=56.2 kJ
The enthalpy of reaction for the reaction:
2H2 (g) + O2(g) $\to $2H2O(l) $\Delta $ rH$\ominus $= - 572 kJmol-1. What will be the standard enthalpy of formation of H2O(l) ?
Ans: Standard molar enthalpy of formation is the enthalpy change for the formation of one mole of a compound from its most stable states or reference states. As per the given information in the question, the standard enthalpy for the given equation is – 572 kJ mol–1
Now the enthalpy of formation for \[H_{2}O\] will be half the enthalpy of the value in the given equation. So, now we can calculate that:
$\Delta $fH=$\frac{~~~1~~}{2}\times \Delta $yH= $\frac{~~~572~~}{2}$=-286KJmol-1