Rate law from data, not from coefficients: three runs give k[NO]^2[H2] in an SC190 Unit 1 reaction rate problem set that adds half-life and activation energy work. Searches like "sc 190 unit 1 assignment example", "sc190 unit 1 sample" and "sc190 unit 1 example" land here.
What a finished SC190 Unit 1 reaction rate problem set looks like
Four problems of rising scope. The first presents initial rates for 2NO plus 2H2 giving N2 plus 2H2O: runs at 0.100 M of each reactant give 1.23 x 10^-3 M/s, doubling NO to 0.200 M gives 4.92 x 10^-3, and doubling H2 instead gives 2.46 x 10^-3. Ratios of runs isolate each exponent, 4 equals 2 to the n for NO and 2 equals 2 to the m for H2, and substituting run one gives k equal to 1.23 M^-2 s^-1. Problem two predicts a rate from that law. Problem three treats a first-order decomposition with k of 5.00 x 10^-4 s^-1, giving a half-life of 1.39 x 10^3 s and 40.7 percent remaining after 30.0 minutes. Problem four turns rate constants at 300 K and 320 K into an activation energy of 55.3 kJ/mol.
How a SC190 Unit 1 example is structured
Each problem restates its question and lists the known quantities, then shows its reasoning before its arithmetic. The orders problem is the centerpiece and is set out as a sequence of run comparisons: two runs where only one concentration changes, their rates divided, the exponent solved and stated as an integer beside the ratio that justifies it. The rate law is written in full before k is calculated, and k is reported with units derived from the overall order. A paragraph then asks whether the law fits a plausible mechanism: a fast equilibrium forming N2O2, followed by a slow step consuming one H2, would produce exactly this dependence. The integrated-rate and Arrhenius problems follow, each with its equation stated, the natural log step shown and the significant figures matched to the data given, three throughout.
Runs compared in pairs
Runs one and two differ only in NO, runs one and three only in H2, and the set names each pair before dividing, so every exponent traces to a specific comparison.
Exponents that ignore coefficients
Hydrogen's coefficient of 2 and its measured order of 1 sit side by side, making the point that an order comes from the experiment and the balanced equation cannot supply it.
Units of k derived, not recalled
A third-order law leaves M^-2 s^-1 on the rate constant, and the set shows the algebra that produces those units from M/s divided by M cubed.
A mechanism that fits
A fast pre-equilibrium between NO and N2O2, then a slow step with H2, reproduces second order in NO and first in H2. The set offers it as consistent with the data, not proven by it.
Logarithms kept to their precision
The half-life, 1.39 x 10^3 s, and the fraction remaining, 0.407, keep three figures from k, and the Arrhenius answer notes that its precision leans on a 20 K temperature difference.
Where marks go in SC190 Unit 1
Rate laws written from the balanced equation, second order in hydrogen here, are the most expensive mistake on this unit because every later answer inherits the wrong exponent. Orders reported without the ratio that produced them come next; a correct 2 with no comparison shown earns little in most sections. Rate constants without units, or with units copied from a different order, lose precision credit. Integrated-rate problems slip when minutes and seconds are mixed, 30.0 minutes used against a k in reciprocal seconds, which turns 40.7 percent remaining into near-total survival. Arrhenius calculations lose marks for temperatures left in Celsius or for R taken as 0.08206 instead of 8.314 J/mol K. Mechanism paragraphs claiming the rate law proves a mechanism, rather than supporting it, draw a reasoning comment in many rubrics.
Get a SC190 Unit 1 example written to your instructions
Attach the Unit 1 data tables and questions verbatim from SC190, noting any required format for rate constants, and include the rubric. Each order is derived from paired runs, the rate law written in full and every log step shown. No fee applies to a first sample, which arrives within 24-48h.
SC190 Unit 1 questions, answered
Why can't the rate law come from the balanced equation?
Because a balanced equation describes the overall change, not the path molecules take to get there. Most reactions proceed through several steps, and the slowest step controls the rate. Only an elementary step's rate law can be read from its coefficients. For an overall reaction, the orders have to come from experimental data such as initial rates.
How are the units of k determined?
From the overall order. Rate is always in M/s, so k must carry whatever units make the right side of the rate law match. A first-order law gives k in reciprocal seconds, second order gives M^-1 s^-1, and third order, as in this set, gives M^-2 s^-1. Writing the algebra once makes the units impossible to misremember.
Does SC190 expect a mechanism for every rate law?
Not always. Many Unit 1 prompts ask only for orders and k. When a mechanism is requested, the expectation is a sequence of steps whose slow step reproduces the measured law, described as consistent with the data. A rate law can rule a mechanism out, but it cannot prove one correct, and graders look for that distinction.