For the reaction 298 K- 2A-B displaystyle rightarrow C displaystyle bigtriangleup H - 400 KJ displaystyle mol-1- and displaystyle bigtriangleup S - 0-02 KJ displaystyle K-1- displaystyle mol-1- At what temperature will the reaction become spontaneous considering displaystyle bigtriangleup H and displaystyle bigtriangleup S to be constant over the temperature range-

Question

For the reaction 298 K- 2A-B  displaystyle rightarrow C  displaystyle bigtriangleup  H - 400 KJ  displaystyle mol-1-  and  displaystyle bigtriangleup  S - 0-02 KJ  displaystyle K-1-   displaystyle mol-1- At what temperature will the reaction become spontaneous considering  displaystyle bigtriangleup  H and  displaystyle bigtriangleup  S to be constant over the temperature range-

Toppr Admin · Accepted Answer

-x394-G-x394-H-x2212-T-x394-SAssuming that equation at equilibrium then-x394-G - 0-x394-G-x394-H-x2212-T-x394-S-0T-x394-H-x394-S-4000-2-2000K-For the reaction to be spontaneous -x394-G must be negative-Therefore temprature should be greater than 2000 K-

For the reaction at 298 K, 2A+B $\to C$

$△$ H = 400 KJ $m o l^{- 1}$ and $△$ S = 0.02 KJ $K^{- 1}$ $m o l^{- 1}$
At what temperature will the reaction become spontaneous considering $△$ H and $△$ S to be constant over the temperature range?

$Δ G = Δ H - T Δ S$

Assuming that equation at equilibrium then

$Δ G$ = $0$

$Δ G = Δ H - T Δ S = 0$

$T = \frac{Δ H}{Δ S} = \frac{400}{0.2} = 2000 K$ .

For the reaction to be spontaneous $Δ G$ must be negative.
Therefore temprature should be greater than 2000 K.

For the reaction at 298 K, 2A+B →C△ H = 400 KJ mol−1 and △ S = 0.02 KJ K−1 mol−1At what temperature will the reaction become spontaneous considering △ H and △ S to be constant over the temperature range?

ΔG=ΔH−TΔSAssuming that equation at equilibrium thenΔG = 0ΔG=ΔH−TΔS=0T=ΔHΔS=4000.2=2000K.For the reaction to be spontaneous ΔG must be negative.Therefore temprature should be greater than 2000 K.

For the reaction at 298 K, 2A+B $\to C$

$△$ H = 400 KJ $m o l^{- 1}$ and $△$ S = 0.02 KJ $K^{- 1}$ $m o l^{- 1}$
At what temperature will the reaction become spontaneous considering $△$ H and $△$ S to be constant over the temperature range?

$Δ G = Δ H - T Δ S$

Assuming that equation at equilibrium then

$Δ G$ = $0$

$Δ G = Δ H - T Δ S = 0$

$T = \frac{Δ H}{Δ S} = \frac{400}{0.2} = 2000 K$ .

For the reaction to be spontaneous $Δ G$ must be negative.
Therefore temprature should be greater than 2000 K.