The reaction CuSO4-3H2Os- Cu-SO4-H2Os-2H2Og- the dissocitaion pressure is 7-10-3atm at 25oC and -Ho-2750cal- What will be the dissociation pressure at 127oC-

K_p_25^∘=7× 10^ 3 atm, T_1=25^∘C=298 k T_2=127^∘C=400k, H^∘=2750 cal lnK_p_27K_p_25= HR[1T_1 1T_2] In, K_p7× 10^ 3=27502[1298 1400] Solving ...

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Standard XII

Physics

Law of Radioactivity

The reaction ...

Question

The reaction $C u {S O}_{4} .3 H_{2} O (s) ⇌ C u . {S O}_{4} . H_{2} O (s) + 2 H_{2} O (g)$ ; the dissocitaion pressure is $7 \times 10^{- 3} a t m$ at $25^{o} C$ and $Δ H^{o} = 2750 c a l$ . What will be the dissociation pressure at $127^{o} C$ ?

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Solution

$K_{p_{25^{\circ}}} = 7 \times 10^{- 3} a t m, T_{1} = 25^{\circ} C = 298 k$
$T_{2} = 127^{\circ} C = 400 k, △ H^{\circ} = 2750 c a l$
$ln \frac{K_{p_{27}}}{K_{p_{25}}} = \frac{△ H}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$
In, $\frac{K_{p}}{7 \times 10^{- 3}} = \frac{2750}{2} [\frac{1}{298} - \frac{1}{400}]$
Solving this, we get
$K_{p} = 104.97 \times 10^{- 3} a t m$

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Similar questions

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$K_{p}$ for the reaction $C u S O_{4} .3 H_{2} O_{(s)} + 2 H_{2} O_{(g)} \to C u S O_{4} . 5 H_{2} O_{(s)} i s 1.0 \times 10^{4} a t m^{- 2}$ at certain temperature. What lowest relative humidity of air can be achieved using $C u S O_{4} .3 H_{2} O$ as drying agent at that temperature ? Aqueous tension at the given temperature = 16 torr

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{N H}_{4} + H^{+} ⇌ {N H}_{4}^{+}; Δ H^{o} = - 52.2 k J {m o l}^{- 1}; Δ S^{o} = + 1.67 J K^{- 1} {m o l}^{- 1}

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The reaction CuSO4.3H2O(s)⇌Cu.SO4.H2O(s)+2H2O(g); the dissocitaion pressure is 7×10−3atm at 25oC and ΔHo=2750cal. What will be the dissociation pressure at 127oC?

Kp25∘=7×10−3atm,T1=25∘C=298kT2=127∘C=400k,△H∘=2750callnKp27Kp25=△HR[1T1−1T2]In, Kp7×10−3=27502[1298−1400]Solving this, we getKp=104.97×10−3atm

The reaction $C u {S O}_{4} .3 H_{2} O (s) ⇌ C u . {S O}_{4} . H_{2} O (s) + 2 H_{2} O (g)$ ; the dissocitaion pressure is $7 \times 10^{- 3} a t m$ at $25^{o} C$ and $Δ H^{o} = 2750 c a l$ . What will be the dissociation pressure at $127^{o} C$ ?