Asked by ally
                Given the following general reaction: 
A + 2B + 3C -> P + 4Q.
Show how the change in concentration of C with time is related to the change in concentration of A, B, P, and Q with time.
            
        A + 2B + 3C -> P + 4Q.
Show how the change in concentration of C with time is related to the change in concentration of A, B, P, and Q with time.
Answers
                    Answered by
            bobpursley
            
    C changes three times as fast as A, 1.5 times as fast as B, and C also changes three times fast as P (but in the opposite direction).   
    
                    Answered by
            Doc48
            
    Remember, these type kinetics problems are always based upon ‘pairs’ of substances. A quick setup is to equate the rates of the two substances of interest and switch the coefficients. One of the rates will be given. Solve for the unknown in terms of the given rate value. For this problem…
A + 2B + 3C => P + 4Q
1 (∆[C])/∆t=3 ([∆A])/([∆t])
2 (∆[C])/∆t=3 ([∆B])/([∆t])
1 (∆[C])/∆t=3 ([∆P])/([∆t])
4 (∆[C])/∆t=3 ([∆Q])/([∆t])
    
A + 2B + 3C => P + 4Q
1 (∆[C])/∆t=3 ([∆A])/([∆t])
2 (∆[C])/∆t=3 ([∆B])/([∆t])
1 (∆[C])/∆t=3 ([∆P])/([∆t])
4 (∆[C])/∆t=3 ([∆Q])/([∆t])
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