In the homogeneous decomposition of nitrous oxide N20 to form nitrogen and oxygen, it is found that at constant temperature the time needed for half the reaction to be completed (T1/2) is inversely proportional to the initial pressure (po) of N2O. On varying the temperature the following results were obtained. Temperature/K967 Po/Torr T1/2 Is 1030 360 212 1085 345 53 294 1520 You should note that 1 atm 760 Torr = 101.325 kPa, and you may assume that N2O behaves as an ideal gas. Deduce the reaction order and calculate: the rate constant k for the reaction at 967 K expressed in units of dm3mol's The mole fraction of nitrogen in the reaction mixture at time t1/2. (ii)

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This should produce a curve with a slope that is equal to the reaction order.

The rate constant k for the reaction at 967 K can then be calculated from the equation:

k = po/t1/2

Using the given data, the rate constant k for the reaction at 967 K can be calculated as:

k = 360/212 = 1.70 dm3mol-1s-1

The mole fraction of nitrogen in the reaction mixture at time t1/2 can be calculated by using the equation:

XN2 = 1/2[1/k(p)t+1]

Using the given data, the mole fraction of nitrogen in the reaction mixture at time t1/2 can be calculated as:

XN2 = 1/2[1/1.70(1030)212+1] = 0.857

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