The reaction: (CH3)3CBr + OH− → (CH3)3COH + Br− in a certain solvent is first order with respect to OH−. In several experiments, the rate constant k was determined at different temperatures. A plot of ln(k) vs 1/T was constructed and the slope of the line was -1.10 x 104 K with a y-intercept of 33.5. What is the activation energy for this reaction? A) -91,400 J/mol
B) -1.10 × 10⁴ J/mol
C) 1.10 × 10⁴ J/mol
D) 91,400 J/mol​

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explanation:

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml

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