GRE Subject Test: Chemistry : GRE Subject Test: Chemistry

Study concepts, example questions & explanations for GRE Subject Test: Chemistry

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Example Questions

Example Question #1 : Reaction Rates And Reaction Order

A scientist is studying a reaction, and places the reactants in a beaker at room temperature. The reaction progresses, and she analyzes the products via NMR. Based on the NMR readout, she determines the reaction proceeds as follows:

\(\displaystyle NH_{4}^{+}(aq) + NO_{2}^{-}(aq) \rightarrow N_{2}(g) +2H_{2}O(l)\)

In an attempt to better understand the reaction process, she varies the concentrations of the reactants and studies how the rate of the reaction changes. The table below shows the reaction concentrations as she makes modifications in three experimental trials.

\(\displaystyle \begin{matrix} Trial & [NH_4^+] & [NO_2^-] &Rate \\ 1& 0.480M &0.120M &0.018\frac{M}{s} \\ 2& 0.240M & 0.120M& 0.009\frac{M}{s}\\ 3& 0.240M& 0.360M & 0.027\frac{M}{s} \end{matrix}\)

 

Which of the following most closely approximates the rate law for this reaction?

Possible Answers:

\(\displaystyle Rate=k[NH_4^+][NO_2^-]^2\)

\(\displaystyle Rate=k[NH_4^+]\)

\(\displaystyle Rate=k[NH_4^+][NO_2^-]\)

\(\displaystyle Rate=k[NH_4^+]^2[NO_2^-]^2\)

\(\displaystyle Rate=k[NH_4^+]^2[NO_2^-]\)

Correct answer:

\(\displaystyle Rate=k[NH_4^+][NO_2^-]\)

Explanation:

The reaction table in the passage indicates that the reaction rate varies in a 1-to-1 fashion as you vary the each reactant, while holding the other constant.

The rate law is written as \(\displaystyle Rate=k[NH_4^+]^x[NO_2^-]^y\).

Compare trials 1 and 2 to see that doubling the ammonium concentration doubles the rate. The reaction is first order for ammonium: \(\displaystyle x=1\).

Compare trials 2 and 3 to see that tripling nitrate concentration triples the rate. The reaction is first order for nitrate: \(\displaystyle y=1\).

The final rate law is \(\displaystyle Rate=k[NH_4^+][NO_2^-]\).

 

Example Question #1 : Reaction Rates And Reaction Order

For the reaction, \(\displaystyle 2X\ +\ Y\rightarrow\ X_{2}Y\), the initial rate of the reaction was determined and the values are tabulated below.

Screen shot 2015 11 27 at 5.48.47 am

What is the overall order for the rate law of this reaction?

Possible Answers:

\(\displaystyle 1\)

\(\displaystyle 2\)

\(\displaystyle 4\)

\(\displaystyle 3\)

Correct answer:

\(\displaystyle 3\)

Explanation:

For the reaction given:

\(\displaystyle rate\ of\ reaction=k[X]^{a}[Y]^{b}\)

We need to find the reaction order, therefore we have to find the values of a and b in the equation. First, let us compare the results of the experiments done to determine how changing the concentration effects the rate of the reaction. Let's compare run number 1 with run number 2 in which the concentration of Y was doubled in run 2 while keeping the concentration of X constant. We can observe that the reaction rate was quadrupled. Let's compare by ratios:

\(\displaystyle \frac{rate_{2}}{rate_{1}}=\left ( \frac{[X]_{2}^{a}[Y]_{2}^{b}}{[X]_{1}^{a}[Y]_{1}^{b}} \right )\)

The value for the concentration of B was the same for run 1 and 2 and therefore they cancel out:

\(\displaystyle \frac{rate_{2}}{rate_{1}}=\left ( \frac{[Y]_{2}^{b}}{[Y]_{1}^{b}} \right )\)

Plugging the values into the equation gives:

\(\displaystyle \frac{3.0\frac{M}{s}}{0.75\frac{M}{s}}=\frac{(0.01M)^{b}}{(0.005M)^{b}}\)

Simplifying the above equation gives:

\(\displaystyle 4.0=2.0^{b}\)

Therefore, \(\displaystyle b=2\)

We can also compare the results of run number 2 and run number 3. In that case, the concentration of Y was kept constant while the concentration of X was doubled. We can see that the reaction rate doubled. Therefore the reaction is first order with respect to X and second order with respect to Y. Therefore, the reaction is third order overall:

\(\displaystyle rate\ of\ reaction=k[X][Y]^{2}\)

Example Question #1 : Reaction Rates And Reaction Order

For the reaction, \(\displaystyle A\ +\ B\rightarrow\ C\ +\ D\), the initial rate of the reaction was determined and the values are tabulated below.

Screen shot 2015 11 27 at 5.58.58 am

Determine the overall order for the rate law of this reaction.

Possible Answers:

\(\displaystyle 3\)

\(\displaystyle 2\)

\(\displaystyle 1\)

\(\displaystyle 0\)

Correct answer:

\(\displaystyle 2\)

Explanation:

For the reaction given:

\(\displaystyle rate\ of\ reaction=k[A]^{x}[B]^{y}\)

We need to find the reaction order, therefore we have to find the values of x and y in the equation. First, let us compare the results of the experiments done to determine how changing the concentration effects the rate of the reaction. Let's compare run number 1 with run number 2 in which the concentration of A was doubled in run 2 while keeping the concentration of B constant. We can observe that the reaction rate was doubled. Let's compare by ratios:

\(\displaystyle \frac{rate_{2}}{rate_{1}}=\left ( \frac{[A]_{2}^{x}[B]_{2}^{y}}{[A]_{1}^{x}[B]_{1}^{y}} \right )\)

The value for the concentration of B was the same for run 1 and 2 and therefore they cancel out:

\(\displaystyle \frac{rate_{2}}{rate_{1}}=\left ( \frac{[A]_{2}^{x}}{[A]_{1}^{x}} \right )\)

Plugging the values into the equation gives:

\(\displaystyle \frac{5.0\times 10^{-5}}{2.0\times 10^{-5}}=\frac{(4.0\times 10^{-3})^{x}}{(2.0\times 10^{-3})^{x}}\)

Simplifying the above equation gives:

\(\displaystyle 2.0=2.0^{x}\)

Therefore, \(\displaystyle x=1\)

We can also compare the results of run number 2 and run number 3. In that case, the concentration of A was kept constant while the concentration of B was decreased by half. We can see that the reaction rate decreased by half. Therefore the reaction is first order with respect to A and first order with respect to B. Therefore, the reaction is second order overall:

\(\displaystyle rate=k[A][B]\)

Example Question #261 : Gre Subject Test: Chemistry

For any given chemical reaction, one can draw an energy diagram. Energy diagrams depict the energy levels of the different steps in a reaction, while also indicating the net change in energy and giving clues to relative reaction rate. 

Below, a reaction diagram is shown for a reaction that a scientist is studying in a lab. A student began the reaction the evening before, but the scientist is unsure as to the type of the reaction. He cannot find the student’s notes, except for the reaction diagram below.

 

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Upon further review, the scientist realizes that the reaction in question involved formation of a carbocation that quickly reacted again to form stable products. At which point would we most likely find this carbocation in the above diagram?

Possible Answers:

5

3

4

1

2

Correct answer:

3

Explanation:

Point 3 is where you would expect to find a relatively stable intermediate. An intermediate is more stable than a transition state, but not as stable as the original reactants and final products. Stability is inversely proportional to energy, thus we are looking for the point that is between the highest and lowest energies in the reaction. By this logic, point 1 is the reactants, 2 and 4 are transition states, 3 is a stable intermediate, and 5 is the products.

Example Question #1 : Reaction Energetics And Kinetmatics

What kinetic equation describes the rate of an E1 process?

Possible Answers:

\(\displaystyle k[\text{R-Br}]\)

None of these

\(\displaystyle k[\text{Base}]\)

\(\displaystyle k[\text{R-Br}]^2\)

\(\displaystyle k[\text{R-X}][\text{Base}]\)

Correct answer:

\(\displaystyle k[\text{R-Br}]\)

Explanation:

Most organic reactions are carried out in multiple steps. The rate equation can be derived from the process that occurs in the slowest step of the mechanism. An E1 reaction's slow step is when a leaving group separates from the hydrocarbon and a carbocation is formed. The only reactant in this step is one molar equivalent of the hydrocarbon. Thus, the rate equation only depends on that substance. The molar equivalent determines the exponent of each reactant in the equation.

Example Question #1 : Analytical Chemistry

Robert conducted an experiment in which he investigated how much water a paper towel could absorb. Initially, Robert found that one paper towel can absorb 12.8g of water. Later he found that his scale was not calibrated, so he had to repeat the experiment. After repeating the experiment with a new scale, Robert found that one paper towel can actually absorb 32.9g of water. What is the approximate percent error between the findings of the first and second experiments?

Possible Answers:

\(\displaystyle 39\%\)

\(\displaystyle 72\%\)

\(\displaystyle 157\%\)

\(\displaystyle 20\%\)

\(\displaystyle 61\%\)

Correct answer:

\(\displaystyle 61\%\)

Explanation:

The formula for percent error is: 

\(\displaystyle \frac{\left | \text{Accepted Value - Measured Value}\right |}{\text{Accepted Value}} \times100\%\)

In this case, the measured value is 12.8g and the accepted value is 32.9g.

\(\displaystyle \frac{\left | 32.9g-12.8g\right |}{32.9g}\times 100\%=61\%\)

Example Question #2 : Precision, Accuracy, And Error

A reaction between one mole of sodium and one mole of chloride should yield 42 grams of sodium chloride. In your experiment, the actual yield is 32.73 grams. Calculate the percent error of your experiment.

Possible Answers:

\(\displaystyle 15.42\%\ \text{error}\)

\(\displaystyle 25.13\%\ \text{error}\)

\(\displaystyle 24.07\%\ \text{error}\)

\(\displaystyle 22.07\%\ \text{error}\)

Correct answer:

\(\displaystyle 22.07\%\ \text{error}\)

Explanation:

To find percent error we need to use the following equation:

\(\displaystyle \frac{\text{theoretical yield}-\text{actual yield}}{\text{theoretical yield}}*100\%\)

Plug in 42 for the theoretical yield and 32.73 for the actual yield and solve accordingly.

\(\displaystyle \frac{42g-32.73g}{42g}*100\%=22.07\%\)

Example Question #1 : Analytical Chemistry

In the following reaction, eight moles of sodium hydroxide is broken down into four moles of sodium oxide and four moles of water. What is the percent error if your experiment yields 195 grams of sodium oxide?

\(\displaystyle 2NaOH\rightarrow Na_{2}O + H_{2}O\)

Possible Answers:

\(\displaystyle 22.15\%\ \text{error}\)

\(\displaystyle 34.17\%\ \text{error}\)

\(\displaystyle 78.63\%\ \text{error}\)

\(\displaystyle 21.37\%\ \text{error}\)

Correct answer:

\(\displaystyle 21.37\%\ \text{error}\)

Explanation:

To find the percent error we need to use the following equation:

\(\displaystyle \frac{\text{theoretical yield}-\text{actual yield}}{\text{theoretical yield}}*100\%\)

But in order to do this, we first have to convert moles of sodium oxide into grams:

\(\displaystyle 4 mol\ Na_{2}O * \frac{62g}{1 mol}=248g\ NaO_2\)

This gives us a theoretical yield of 248g, which we plug in with our 195g actual yield.

\(\displaystyle \frac{248g-195g}{248g}*100\%=21.37\%\ \text{error}\)

Example Question #61 : Measurements

If a given sample of silver and flourine ideally combine to form 0.6mol of AgF, what is the percent error if the actual yield is 43 grams?

Possible Answers:

\(\displaystyle 51.98\%\ \text{error}\)

\(\displaystyle 47.81\%\ \text{error}\)

\(\displaystyle 56.43\%\ \text{error}\)

\(\displaystyle 43.57\%\ \text{error}\)

Correct answer:

\(\displaystyle 43.57\%\ \text{error}\)

Explanation:

Our first step to complete this problem is to convert moles AgF into grams:

\(\displaystyle 0.6 mol\ AgF*\frac{127g}{1 mol}=76.2g\ AgF\)

This gives us a theoretical yield of 76.2 grams. We can find the percent error by plugging in 76.2g for our theoretical yield and 43g for the actual yield in the following equation:

\(\displaystyle \frac{\text{theoretical yield}-\text{actual yield}}{\text{theoretical yield}}*100\%\)

\(\displaystyle \frac{76.2g-43g}{76.2g}*100\%=43.57\%\ \text{error}\)

Example Question #261 : Gre Subject Test: Chemistry

Michael buys several bags of balloons. On the package, it says that each bag has 100 balloons. He opens the bags and only one of them has 100 balloons inside; the other bags either have too many or too few.

How would you describe the bag of balloons with 100 balloons inside?

Possible Answers:

Both accurate and precise

Accurate, but not precise

Accurate, but the precision cannot be determined

Neither accurate nor precise

Precise, but not accurate

Correct answer:

Accurate, but not precise

Explanation:

This bag is accurate because it provided the correct number of balloons, however, the process is not precise as the results were clearly not repeatable.

Accuracy deals with how close the measurement got to the accepted measurement. Precision deals with how consistent the measurement is. The bag with 100 balloons inside matched the claim made on the bag, meaning it was accurate. It was not precise because the other measurements show that the number of balloons is variable.

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