`A to A, DeltaH=+Ve`, Graph between `log_(10)A and 1/T` is straight line of slpe `1/(4.606)` find `DeltaH` in cel

A rarr B , Graph between long_(10) P and (1)/(T) is a straight line of slope (1)/(4.606) . Hence, Delta H is

The graph between log k versus 1//T is a straight line.

The graph between log k versus 1//T is a straight line.

The graph between the log K versus 1/T is a straight line. The slope of the line is

The graph between. The log K verrus (1)/(T) is a straight line. The slope of the line is

In the graph drawn between log K and 1/T, intercept equalst o

A graph plotted between log_(10) K_(c) and 1//T is straight line with intercept 10 and slpoe equal t0 0.5. Calculate : ( i ) pre -exponential factor A . ( ii ) heat of reaction at 298 K . ( iii ) equilibrium constant at 298 K . ( iv ) equilibrium constant in at 800 k assuming DeltaH remains constant in between 298 K and 800 K .

Variation of equiibrium constant K with temperature T is given by van 't Holf equation, log K=logA-(DeltaH^(@))/(2.303RT) A graph between log K and T^(-1) was a straight line as shown in the figure and having theta=tan^(-1)(0.5)"and"OP=10 . Calculate: (a) DeltaH^(@) (standard heat of reaction)when T=300K, (b) A (pre-exponetial factor), (c) Equilibrium constant K at 300 K, (d) K "at" 900 K "if" DeltaH^(@) is independent of temperature.