ΔS and \( \Delta H\) are having negative sign.
The reaction given in the question represents the formation of Cl molecule from Cl atoms. As the formation of bond takes place in the given reaction. So, energy is released. So,\( \Delta H\) is negative.
Also, 2 moles of Chlorine atoms are having more randomness than 1 mole of chlorine molecule. So, the spontaneity is decreased. Thus, \( \Delta S\) is negative.
Answered by Pragya Singh | 1 year agoCalculate the entropy change in surroundings when 1.00 mol of \( H_2O\)(l) is formed under standard conditions. \( Δ_f H^ Θ= –286 kJ mol^{–1}.\)
Comment on the thermodynamic stability of NO(g), given,
(\( \dfrac{1}{2}\))N2(g) + (\( \dfrac{1}{2}\))O2(g) → NO(g); \( \Delta _{r}H^{\Theta } = 90kJmol^{-1}\)
NO(g) + (\( \dfrac{1}{2}\))O2(g) → NO2(g); \( \Delta _{r}H^{\Theta } = -74kJmol^{-1}\)
The equilibrium constant for a reaction is 10. What will be the value of ∆G0? R = 8.314 JK–1 mol–1, T = 300 K.
For the reaction
2A(g) + B(g)→ 2D(g)
\( \Delta U^{\Theta } = -10.5 kJ \;and \;\Delta S^{\Theta } = -44.1JK^{-1}\)
Calculate \( \Delta G^{\Theta }\) for the reaction, and predict whether the reaction may occur spontaneously.
For the reaction at 298K,
2A + B → C
\( \Delta H = 400 kJ mol^{-1}\)
\( \Delta H = 0.2 kJ mol^{-1}K^{-1}\)
At what temperature will the reaction become spontaneous considering \( \Delta S\; and\; \Delta H\) to be constant over the temperature range?