As an experienced academic specialist at LiveExamHelper.com, I regularly support students tackling high-pressure, timed assessments. If you’ve ever found yourself wondering, “Can someone Take My Proctored Chemistry Exam For Me?”—you’re not alone. Many graduate-level students juggling jobs, internships, and personal commitments turn to us for assistance. We provide qualified subject-matter experts who are proficient in proctored testing environments and well-versed in upper-level chemistry topics.

To give you a glimpse into the caliber of work our experts deliver, below are two master’s-level chemistry exam questions recently completed for a client. These questions were taken from a graduate thermodynamics and kinetics module and highlight the analytical depth and precision we offer in both live exam and take-home scenarios.

Sample Question One: Thermodynamics – Gibbs Free Energy and Equilibrium

Question:
A reaction is carried out at 298 K where ΔH° = -92.0 kJ/mol and ΔS° = -198 J/mol·K. Calculate ΔG° for the reaction. Is the reaction spontaneous under standard conditions?

Solution by our expert:
To calculate ΔG°, we use the standard Gibbs Free Energy formula:

ΔG° = ΔH° - TΔS°

Before substituting the values, it’s important to ensure consistent units. ΔH° is in kJ/mol, so we must convert ΔS° into kJ/mol·K:

ΔS° = -198 J/mol·K = -0.198 kJ/mol·K

Now substitute the values:

ΔG° = -92.0 kJ/mol - (298 K)(-0.198 kJ/mol·K)
ΔG° = -92.0 kJ/mol + 59.004 kJ/mol
ΔG° = -32.996 kJ/mol

Since ΔG° is negative, the reaction is spontaneous under standard conditions.

Analysis:
This problem integrates fundamental concepts of thermodynamics with unit conversion, which is often a stumbling block under timed conditions. Our expert not only solved the question but also clarified unit handling—something that impresses instructors and maintains academic integrity.

Sample Question Two: Kinetics – Reaction Mechanisms and Rate Laws

Question:
For the following reaction:
2 NO₂(g) → 2 NO(g) + O₂(g)

The proposed mechanism involves:
Step 1 (slow): NO₂ + NO₂ → NO₃ + NO
Step 2 (fast): NO₃ → NO + O₂

Derive the rate law based on the mechanism and identify the rate-determining step.

Solution by our expert:
The rate-determining step is the slow step, which dictates the rate law.

Step 1: NO₂ + NO₂ → NO₃ + NO (slow)
This is bimolecular, involving two NO₂ molecules.

Therefore, the rate law is:
Rate = k[NO₂]²

Analysis:
The student had initially misidentified the rate-determining step as Step 2 due to its simplicity. Our expert provided a thorough explanation, pointing out that only the slow step contributes to the rate law and clarified the relationship between molecularity and reaction order.

How We Can Help You Succeed

Whether you're preparing for an open-book graduate exam or struggling with a proctored online assessment, we match you with professionals who can handle exams in real-time or under academic supervision. All our solutions are crafted to meet university-level standards in accuracy, clarity, and formatting. Our chemistry experts have deep experience in thermodynamics, kinetics, quantum chemistry, and more.

If your concern is “Can I trust someone to Take My Proctored Chemistry Exam For Me and deliver accurate, timely results?”—the answer is yes, with LiveExamHelper.com. We offer secure, discreet services that have helped hundreds of graduate students perform better without sacrificing their other responsibilities.

Reach out today to connect with a chemistry expert who can not only take the pressure off your exam prep but ensure results that reflect mastery and professionalism.