[" 33"34" 35"36" 37"38],[" For a reversible spontaneous change,"Delta" S is "]

In a reversible adiabatic change Delta Q is

In a reversible adiabatic change Delta Q is

In a reversible process, Delta S_(sys) + Delta S_(surr) is

For an exothermic reaction to be spontaneous ( Delta S = negative)

For an exothermic reaction to be spontaneous ( Delta S = negative)

Statement-1: For a process to be spontaneous, Delta G as well as Delta S has to be less than zero. Statement-2 : For spontaneous change, Delta S_("total") gt 0

Explain change in entropy of a system during a reversible process Delta S = (q_(rev))/(T)

For a given reaction , Delta H = 3.5 kJ mol^(-1) and Delta S = 83.6 J * K^(-1) mol^(-1) . The reaction is spontaneous at (assume DeltaH and Delta S do not vary with temperature) -

The weights (in kg) of 30 students of a class are 39,43,32,37,29,26,31,45, 46,31,37,38,30,39,36,41,35,34,41,46,39,38,36,38,40,42,33,43,44,33. Prepare a frequency distribution table using one class interval as 30-35 in which 30 is included and 35 excluded. using the above data, draw a histogram.