Explanation
Heat evolved in the 1st case = 400 × T1 Heat evolved in the 2nd case = Half of 1st case (amounts are half ) = 200 T1 Actual heat evolved = (100 + 100) T2. = 200 T2 Hence T1 =T2.
q2 is heat of neutralization of strong acid with strong base and q1 is that of weak acid HA with strong base. The difference gives the heat of dissocitation of HA, i,e., ΔH = ΔH2 − ΔH1 = (-q) - (-q1) = q1 - q2
As dew formation is spontaneous process therefore entropy or randomness of the universe will increase. As randomeness of the system has decreased but randomness of the surrounding willincrease larger so that change is positive.
It I is because of the fact that for spontaneity, the value of ∆G = (∆H – T∆S) should be < 0. If DS is – ve, the value of T∆S shall have to be less than ∆H or the value of ∆S has to be less than that of x
In an isolated system, there is no exchange of energy or matter between the system and surrounding. For a spontaneous process in an isolated system, the change in entropy is positive, i.e. ∆S > 0
∆H = ∆E + ∆ng RT
Here, ∆ng = 6 – 7.5 = – 1.5
Thus, ∆E = ∆H + ∆ngRT = –780980 – (–1.5) × 2 × 298
= –780090 calories
F2O (g) + H2O(g) → O2(g) + 2HF (g) ; ∆H = 76.1 Kcal.
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