Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimen
tal data provided.
Q + X yields products
Trial [Q] [X] Rate
1 0.12 M 0.10 M 1.5 × 10-3 M/min
2 0.24 M 0.10 M 3.0 × 10-3 M/min
3 0.12 M 0.20 M 12.0 × 10-3 M/min
Q + X yields products
Trial [Q] [X] Rate
1 0.12 M 0.10 M 1.5 × 10-3 M/min
2 0.24 M 0.10 M 3.0 × 10-3 M/min
3 0.12 M 0.20 M 12.0 × 10-3 M/min
1 answer:
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Answer : The heat of combustion per mole of caffeine at constant volume is 76197.18 kJ/mole
Explanation :
First we have to calculate the specific heat calorimeter.
Formula used :
where,
Q = heat of combustion of benzoic acid = 26.38 kJ/g = 26380 J/g
m = mass of benzoic acid = 0.235 g
c = specific heat of calorimeter = ?
= change in temperature =
Now put all the given value in the above formula, we get:
Thus, the specific heat of calorimeter is
Now we have to calculate the heat of combustion of caffeine.
Formula used :
where,
Q = heat of combustion of caffeine = ?
c = specific heat of calorimeter =
= change in temperature =
Now put all the given value in the above formula, we get:
Now we have to calculate the moles of caffeine.
Mass of caffeine = 0.275 g
Molar mass of caffeine = 194.19 g/mole
Now we have to calculate the heat of combustion per mole of caffeine at constant volume.
Therefore, the heat of combustion per mole of caffeine at constant volume is 76197.18 kJ/mole