Question

26 Consider the reaction Data A s products T K (51) 275 0.370 825 O. 782 what two porate Should be plotted to grupically Determine the Errors, use at least three significant figures in values. XI ? x2 = 3 ? /2 = 2 Determine the rise, run and slope of the line formed by these points vise=? to run = ? Slope = ? What is the activation Energy of this reaction? Eq=?

Answer 1

Answer :

Arrhenius equation relate the rate constant (K) , activation energy (Ea) and temperature (T) as

K = A * e^( -Ea / RT)

Taking natural logarithm on both sides

ln K = ln A - EA/ RT

If graph is plotted between ln K and 1/T ,

Slope = -Ea/R

We can calculate activation energy by determining the slope .

T(K)	K	1/T	ln K
275	0.370	0.00364	-0.994
825	0.782	0.00121	-0.246

2 points to be plotted on graph are

On X axis 1/T values are used and on y axis ln K values are used

Y1 = -0.246 , Y2 = -0.994

X1 = 0.00121 , X2 = 0.00364

Run = X2 - X1 = 0.00364 - 0.00121 = 0.00243

Rise = Y2 - Y1 = -0.994 - (-0.246) = - 0.748

Slope = rise / run = -0.748 / 0.00243 = -307.3189

Slope = -Ea / R

R (gas constant) = 8.314 J / mol-K

Ea ( activation energy) = - slope * R = - (-307.8189) * 8.314 = 2559.2 J/mol

= 2.56 kJ /mol

@ y = a + b*x No Weighting -0.2 0 Equation Weight Residual Sum of Squares Pearson's Adj. R-Square -1 Standard Erro -0.4 Value Intercept 0.12646 Slope -307.8189 B In K -0.6 -0.8 -1.0 T 0.001 0.002 0.003 0.004 1/T (K)