I can't get my table to line up so I've surrounded the values in curly brackets.
Experiment [SO2]i [O2]i IRD O2
{1} {0.05} {0.05} {7.3*[10^(-2)]}
{2} {0.10} {0.05} {2.9*[10^(-2)]}
{3} {0.10} {0.15} {2.6*[10^(-1)]}
{4} {0.12} {0.12} {?????????????}
so overall oreder is 4 or specifically:
rate=k([SO2]^2)([O2]^2) with k= 1168 so
rate=1168([SO2]^2)([O2]^2)
not sure what to do here I believe that rate relationship is:
-(delta [O2)]/delta time)=
1/2(delta [SO3]/delta time)
and I think I need to use trials 2&3 because that is where [SO2] is held constant.
Answer is given as 4.84*10^(-1) trying to figure out how to get there.
Thank you
To solve for the value of [O2] in trial 4, we can use the rate equation and the relationship you mentioned:
rate = k([SO2]^2)([O2]^2)
But since we are given the rate constant k and all the other values in trial 4, except [O2], we can rearrange the equation to solve for [O2]:
[O2]^2 = rate / (k * [SO2]^2)
Now, let's substitute the values for rate, k, and [SO2] from trial 4 into the equation:
[O2]^2 = (4.84 * 10^(-1)) / (1168 * (0.12)^2)
Simplifying this equation gives us:
[O2]^2 = 3.409 * 10^(-5)
To find the value of [O2], take the square root of both sides:
[O2] = sqrt(3.409 * 10^(-5))
Evaluating this expression results in:
[O2] = 5.84 * 10^(-3)
Therefore, the value of [O2] in trial 4 is approximately 5.84 * 10^(-3).