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Standard XII

Chemistry

Heat of Formation

ΔrH and ΔrS...

Question

$Δ_{r} H$ and $Δ_{r} S$ for the reaction $M g O (s) + C (s) \to M g (s) + C O (g)$
are $491.18$ kJ $m o l^{- 1}$ and $197.67$ J $m o l^{- 1}$ respectively. Calculate the temperature at which Gibb's energy change will be zero.

2.4 K

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12.4 L

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312.4 K

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12.4 K

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Solution

The correct option is B 2.4 K
The given facts are :-
$M_{g} O (s) + C (s) \to M g (s) + C O (g)$
$△_{r} H = 491.18 K J / m o l, △ r s = 197.67 J / m o l$
Now, we have, $△ G = △ H - T △ S$
for $△ G = 0 \Rightarrow 0 = △ H - T △ S \Rightarrow T = \frac{△ H}{△ S} = \frac{491.18}{197.67} = 2.4 K$

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Similar questions

Q. The reaction,

M g O (s) + C (s) \to M g (s) + C O (g),

For which

△_{r} H^{o} = + 491.1 k J m o l^{- 1} a n d △_{r} S^{\circ} = 198.0 J K^{- 1} m o l^{- 1}

is not feasible at 298 K. Temperature above which reaction will be feasible is

△ H a n d △ S

for the reaction,

A g_{2} O (s) \to 2 A g (s) + \frac{1}{2} O_{2} (g)

are

30.56 k J m o l^{- 1}

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66.0 J m o l^{- 1} K^{- 1}

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$△ H$ and $△ S$ for the reaction: $A g_{2} O (s) \to 2 A g (s) + \frac{1}{2} O_{2} (g)$ are 30.56 kJ $m o l^{- 1}$ and 66.0 J $K^{- 1} m o l^{- 1}$ respectively. Calculate the temperature at which free energy change for the reaction will be zero. Predict whether the forward reaction will be favored above or below this temperature.

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ΔrH and ΔrS for the reaction MgO(s)+C(s)→Mg(s)+CO(g)are 491.18 kJ mol−1 and 197.67 J mol−1 respectively. Calculate the temperature at which Gibb's energy change will be zero.

The correct option is B 2.4 KThe given facts are :- MgO(s)+C(s)→Mg(s)+CO(g)△rH=491.18KJ/mol,△rs=197.67J/molNow, we have, △G=△H−T△Sfor △G=0⇒0=△H−T△S⇒T=△H△S=491.18197.67=2.4K

$Δ_{r} H$ and $Δ_{r} S$ for the reaction $M g O (s) + C (s) \to M g (s) + C O (g)$
are $491.18$ kJ $m o l^{- 1}$ and $197.67$ J $m o l^{- 1}$ respectively. Calculate the temperature at which Gibb's energy change will be zero.