It is found that the equilibrium constant increases by a factor of four when the temperature is increased from 27- C to 47- C- The value of - H - is-Take ln 2-0-7- log 2-0-3- R -25-3 J mol -1 K -1A- 28 kJ - molB- 24 kJ - molC- 12 kJ - molD- 56 kJ - mol

The correct option is D 56 kJ/molUsing the equation, ln(K_p)_47 ^∘C(K_p)_27 ^∘C=Δ HR [ 1T_1 1T_2 ] We get, ln 4 = Δ H25/3 ( 1273+27 1273+47 ) ∴Δ H = 56 kJ/mo ...

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Standard XII

Chemistry

Ideal Gas Equation

It is found t...

Question

It is found that the equilibrium constant increases by a factor of four when the temperature is increased from $27^{\circ} C$ to $47^{\circ} C$ . The value of $Δ H^{\circ}$ is:
Take $l n 2 = 0.7, l o g 2 = 0.3, R = \frac{25}{3} J m o l^{- 1} K^{- 1}$

28 kJ/mol

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12 kJ/mol

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24 kJ/mol

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56 kJ/mol

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Solution

The correct option is D $56 kJ/mol$
Using the equation,
$ln \frac{(K_{p})_{47^{\circ} C}}{(K_{p})_{27^{\circ} C}} = \frac{Δ H}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$

We get,
$ln 4 = \frac{Δ H}{25 / 3} (\frac{1}{273 + 27} - \frac{1}{273 + 47})$

$∴ Δ H = 56 kJ/mol$

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It is found that the equilibrium constant increases by a factor of four when the temperature is increased from 27 ∘C to 47 ∘C. The value of ΔH∘ is: Take ln2=0.7, log2=0.3, R=253 J mol−1 K−1

The correct option is D 56 kJ/molUsing the equation, ln(Kp)47 ∘C(Kp)27 ∘C=ΔHR[1T1−1T2] We get, ln 4=ΔH25/3(1273+27−1273+47) ∴ΔH=56 kJ/mol

It is found that the equilibrium constant increases by a factor of four when the temperature is increased from $27^{\circ} C$ to $47^{\circ} C$ . The value of $Δ H^{\circ}$ is:
Take $l n 2 = 0.7, l o g 2 = 0.3, R = \frac{25}{3} J m o l^{- 1} K^{- 1}$

The correct option is D $56 kJ/mol$
Using the equation,
$ln \frac{(K_{p})_{47^{\circ} C}}{(K_{p})_{27^{\circ} C}} = \frac{Δ H}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$

We get,
$ln 4 = \frac{Δ H}{25 / 3} (\frac{1}{273 + 27} - \frac{1}{273 + 47})$

$∴ Δ H = 56 kJ/mol$