A certain reaction follows zero-order kinetics. Suppose the reaction went to $50%$ completion after $10$ minutes. What percentage of completion does the reaction reach in an additional $5$ minutes?

Start by writing the balanced chemical equation that describes this $color(red)(2)"NO"_ (2(g)) + "F"_ (2(g)) -> 2"NO"_ 2"F"_((g))$ Notice that the reaction consumes $color(red)(2$ moles of nitrogen...

Step 1: Identify the limiting reactant We can use partial pressures as a replacement for moles because $P ∝ n$. $color(white)(mmmmmm)"N"_2color(white)(m) + color(white)(m)"3H"_2 → "2NH"_3$ $P"(atm):"color(white)(mll) 2.73color(white)(mmm)2.73$ $"Divide by:" color(white)(mll)1color(white)(mmmml)3$...

$t_"1/2" = "24.1 s"$. Interesting question. The rates $r(t)$ defined at each point in a reaction shall be given by the first order : $r(10) =...

I got about $"78 s"$. $A)$ is incorrect because it is merely one over $k$. $B)$ is incorrect because it results from taking $24%$ completion to mean $24%$ of reactant...

You will need to be able to write a general and solve for the units... $r(t) = k[A]^n$ $r(t)$ is the initial rate as a function of time $t$,...

Highlight to reveal answers once you've figured it out. $color(white)(0, 2, 1/2, 1, 0)$ $1)$ Well, consider the following half-lives: $t_"1/2" = ([A]_0)/(2k)$ (zero...

The integrated for a first-order reaction looks like this $color(blue)(ln( A/A_0) = - k * t)" "$, where$ $A$ - the concentration at a given time $t$...

The equation given to you looks like this $color(blue)(|bar(ul(color(white)(a/a)k = 2.303/tlog[ (P_0(n-1))/(nP_0-P_t)]color(white)(a/a)|)))$ To make things more general, I'll use concentration instead of pressure. Keep in mind that...

Initial mass $M_0 = 372$ grams Final mass $M_f = 23.25$ grams. Exponential decay says (k is decay constant and t is time) $M_f = M_0timese^(-ktimest)$ $ln(M_f/M_0) = -ktimest$ $ln(23.25/372)...

Recall that the formula for half-life is: $color(blue)(|bar(ul(color(white)(a/a)y=a(b)^(t/h)color(white)(a/a)|)))$ where: $y=$final amount $a=$inital amount $b=$growth/decay $t=$time elapsed $h=$half-life To find the amount of the sample left after $60$ minutes,...