In a certain chemical reaction, a compound A is formed from a compound B. The masses of A and B at time t after the start of the reaction are x and y respectively and the sum of the masses is equal to 50 throughout the reaction. At any time the rate of increase of the mass of A is proportional to the mass of B at that time.
- (i)Explain why dx/dt = k(50 – x), where k is a constant.[1]
- (ii)It is given that x = 0 when t = 0, and x = 25 when t = 10. Solve the differential equation in part (i) and express x in terms of t.[8]
Mark scheme
Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.
- [B1] Justify the given differential equation
- [B1] Separate variables correctly and attempt to integrate one side
- [B1] Obtain term kt, or equivalent
- [B1] Obtain term –ln(50-x), or equivalent
- [M1*] Evaluate a constant, or use limits x = 0, t = 0 in a solution containing terms aln(50-x) and bt
- [A1] Obtain solution –ln(50-x) = kt - ln 50, or equivalent
- [DM1] Use x = 25, t = 10 to determine k
- [A1] Obtain correct solution in any form, e.g. ln 50 - ln(50-x) = (ln 2)t
- [A1] Obtain answer x = 50(1-exp(-0.0693t)), or equivalent
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