[sundrops]

2HI(g)  -> H2(g) + I2(g)
Rate = -d[HI]/dt = k = 1.90×10-4 mol L-1 s-1

If the experiment has an initial HI concentration of 0.280 mol/L, what is the concentration of HI after 15.0 minutes?

ok, so the way I went about solving this problem is this:

rate= - (change in concentration) / (change in time)

so,
1.90*10^-1 mol/L/s = ([HI] - 0.280mol/L)/900s
0.171mol/L = [HI] -0.280mol/L
0.451 mol/L = [HI]

but that doesn;t work. can someone point out where I went wrong please?

[Mitch]

your forgetting a squared term.

[Donaldson Tan]

rate = k [[HI]]²

[charco]

substituting the initial value into the rate expression would give...

rate = k[HI]^x

rate = 1.90×10-4 mol L-1 s-1 = k[0.280 mol/L,]^x

You have two unknowns here.
How are you supposed to get the value of the order x from the data given?