Question

2Na₃PO₄(aq) + 3SrCl₂(aq) → Sr₃(PO₄)₂(s) + 6NaCl(aq)
The concentration of SrCl₂ is 0.744 M at the start of the reaction, and 0.647 M after 136 seconds. The initial concentrations of the products are zero.

1. What is the average rate of reaction (in M/min) over this time period?

2. What is the average rate of change (in M/min) of SrCl₂ in the first 136 seconds.

3. What is the average rate of change (in M/min) of NaCl in the first 136 seconds.

4. What is the concentration (in M) of NaCl after 136 seconds.

Answer 1

1.

average rate = 1/3 rate of that of SrCl2

rate of SrCl2 = (0.647-0.744)/(136) = -0.000713 M/s (negative since it is a reacatnat)

then

average rate = -1/3*-0.000713 = 0.00023766 M/s

2.

the rate for SrCl2 has been already calcualted

rate of SrCl2 = (0.647-0.744)/(136) = -0.000713 M/s (negative since it is a reacatnat)

3.

for Nacl; ratio is 3:6; that is, double for production

rate of Nacl = +6/2*0.000713 = v

4.

the final concnetration o fNaCl must be:

6/3*(0.744-0.647) = 0.194 M