For oxidation of Fe
`4Fe(s) + 3O_2(g) to 2Fe_2O_3(s)`
entropy change is `-549.4 JK^-1mol^-1` at 298K. Inspite of negative entropy change of this reaction, why is the reaction spontaneous ?
`triangleH^@ = -1648xx10^3 J mol^-1`

For the reaction 2A(g) + B(g) to 2D(g), triangleU^@ = -10.5kJ and triangleS^@ = -44.1 JK^-1 Calculate triangleG^@ for the reaction, and predict whetehr the reaction may occur spontaneously. ( R = 8.314 xx 10^-3 kJ K^-1 mol^-1, T = 298 K )

At 1 atm and 298 K, entropy change of the reaction, 4Fe(s)+3O_(2)(g)to2Fe_(2)O_(3)(s) is -549.4 J*K^(_1) . In this reaction, if DeltaH=-1648kJ , then predict whether the reaction is spontaneous or not.

At 1 atm and 298K DeltaH^(0) value of the reaction 2H_(2)(g)+O_(2)(g)to2H_(2)O(l) is -572 kJ. Calculate the change in entropy of the system and surroundings for this reaction. Is this reaction spontaneous at that temperature and pressure? Given: Standard molar entropies of H_(2)(g),O_(2)(g)&H_(2)O(l) at 298K are 130.6, 205.0 and 69.90 J*K^(-1)*mol^(-1) respectively.

The temperature at which the reaction will be spontaneous C + O_2 = CO_2 , Delta H = 170 kJ mol^-1 , DeltaS = 170 J K^(-1) mol^(-1)

Predict the sign of entropy change is the following reactions CaC_2O_4(s)rarr CaCO_3(s)+CO(g)

For a particular reaction, triangleH^@ = -38.3 kJ mol^-1 and triangleS^@ = -113JK^-1 mol^-1 . This reaction is

Predict the sign of entropy change is the following reactions O_2(g)+2SO_2(g)rarr 2SO_3(g)

In the reaction 1/2N_(2(g)) + 3/2 H_(2(g)) to NH_(3(g)) . The standard entropies of N_(2(g)), H_(2(g)) and NH_(3(g)) are 191.6,130.5 and 192.5 JK^(-1) "mol"^(-1) respectivly. If free energy charge of the reaction is -16.67 kJ. Calculate the Delta H_("reaction")^@ for the formation of NH_3 at 298 K.

Predict the sign of entropy change is the following reactions 2H_2(g)+O_2(g)rarr 2H_2O(g)