Explanation
ΔE is a state function but w and q are not state functions.
This is because electron is attracted by the nucleus.
In (b) solid changes into gas . Hence ΔS is maximum
Dilution of conc. H2SO4 is an exothermic process.
ΔG = ΔH – TΔS. For ΔG to be negative, TΔS > ΔH i.e, 300 ΔS > 15000 J or ΔS > 50J.
+ve ΔH and –ve ΔS both oppose the reaction.
Heat of neutralisation of strong acid with strong base is – 57 kJ mol-1.
As heat is evolved, heat content of diamond > heat content of graphite.Hence graphite is more stable.
N2 + O2 → 2 NO is anendothermic reaction
As the base is weak, heat of neutralisation < 13.7 kcal.
When H2O (g) condenses to form H2O(l), heat is evolved. Hence ΔH2 > ΔH1 or ΔH1 < ΔH2.
In (a), g → l, In (b), np = nr, In (d) np < nr. In (c), s → g. Hence entropy increases.
Heat of formation is for the formation of one mole of the substance fromits elements.
ΔH = –ve and ΔS = +ve both favour the process.
By definition, Hof is for formation of one mole of the compound from its elements in the standard state.
HCl and KOH both are strong.
For the given reaction. Δng = 8 – 8 = 0 Hence ΔH = ΔE.
G = H – TS.
ΔHneut cannot be predicted unless we know heat of dissociation of aceticacid.
All values are equivalent.
For the formation of 1 mole of HCl from elements, ΔHof = ‒22 kcal.
Heat capacity = dQ/dT.
H = E + PV.
When gas is compressed its entropy decreases i,e. ΔS is negative
For S + O2 → SO2, operate eqn. (i) - eqn (ii) .This gives ΔH = y – 2x.
For endothermic reaction, ΔH = +ve. As ΔG = ΔH ‒ TΔS, ΔG is +ve at low temperature if ΔS is +ve but TΔS < ΔH. At high temperature TΔS > ΔH and ΔG becomes ‒ ve.
For (c) , ΔSo > 0 because ions in the aqueous solution have greater randomness than the solid.
Since internal energy is a state function and the system returns tooriginal state, DETAE = 0 (i,e. for a cyclic process ΔE = 0)
For irreversible process (spontaneous process) (dS)V,E > 0’ (dG)T,P < 0
ΔHof is the enthalpy change when 1 mole of the substance is formed from its elements in the standard state.Reaction (a) does not represent ΔHof because standard state of carbon is graphite and not diamond.
Eqn. (i) – Eqn. (ii) gives the required result.
More cis-pentene is formed to restore the equilibrium.
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