Which statement best describes why rate laws are important?

Consider the following reaction. SO2Cl2 → SO2 + Cl2. After collecting experimental data you found that plotting ln[SO2Cl2] vs. t

Nikitich [7]

Answer:

$[SO_2Cl_2]_{600}= 0.0842 M$

Explanation:

Some theoretical knowledge is required here. We should understand that whenever we plot the natural logarithm, ln, of a concentration vs. time and obtain a straight line, this indicates a first-order reaction. That said, since this is the case here, we have a first-order reaction with respect to $SO_2Cl_2$ .

The linear equation has the following terms:

$y = -0.000290t - 2.30$

It is a linear form of the integrated first-order law equation:

$ln[SO_2Cl_2]_t = -kt + ln[SO_2Cl_2]_o$

Therefore, the rate constant, k, is:

$k = 0.000290 s^{-1}$

The natural logarithm of initial molarity is:

$ln[SO_2Cl_2]_o = -2.30$

Using the equation, we may substitute for t = 600 s and obtain the natural logarithm of the concentration at that time:

$ln[SO_2Cl_2]_{600} = -0.000290 s^{-1}\cdot 600 s - 2.30 = -2.474$

Take the antilog of both sides to find the actual molarity:

$[SO_2Cl_2]_{600}=e^{-2.474} = 0.0842 M$

4 0

2 years ago

A friend tells you he read on a web site that earth is getting smaller, What can you tell him that shows Earth's size is not cha

Lyrx [107]

U can show him that Earth isn't getting smaller, because things WOULD be getting smaller, and our houses would be getting smaller, IF everything shrunk WITH Earth.

3 0

3 years ago

Thousand of molecules are called

kozerog [31]

Answer:

monomers

Explanation:

Thousands of molecules are called monomers

6 0

3 years ago

What does the oxidizing agent do in a redox reaction apex?

densk [106]

Same as balancing a regular chemical reaction! Please see the related question to the bottom of this answer for how to balance a normal chemical reaction. This is for oxidation-reduction, or redox reactions ONLY! These instructions are for how to balance a reduction-oxidation, or redox reaction in aqueous solution, for both acidic and basic solution. Just follow these steps! I will illustrate each step with an example. The example will be the dissolution of copper(II) sulfide in aqueous nitric acid, shown in the following unbalanced reaction: CuS (s) + NO 3 - (aq) ---> Cu 2+ (aq) + SO 4 2- (aq) + NO (g) Step 1: Write two unbalanced half-reactions, one for the species that is being oxidized and its product, and one for the species that is reduced and its product. Here is the unbalanced half-reaction involving CuS: CuS (s) ---> Cu 2+ (aq) + SO 4 2- (aq) And the unbalanced half-reaction for NO 3 - is: NO 3 - (aq) --> NO (g) Step 2: Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-reaction. In this case, copper, sulfur, and nitrogen are already balanced in the two half-reaction, so this step is already done here. Step 3: Balance oxygen by adding H 2 O to one side of each half-reaction. CuS + 4 H 2 O ---> Cu 2+ + SO 4 2- NO 3 - --> NO + 2 H 2 O Step 4: Balance hydrogen atoms. This is done differently for acidic versus basic solutions. . For acidic solutions: Add H 3 O + to each side of each half-reaction that is "deficient" in hydrogen (the side that has fewer H's) and add an equal amount of H 2 O to the other side. For basic solutions: add H 2 O to the side of the half-reaction that is "deficient" in hydrogen and add an equal amount of OH - to the other side. Note that this step does not disrupt the oxygen balance from Step 3. In the example here, it is in acidic solution, and so we have: CuS + 12 H 2 O ---> Cu 2+ + SO 4 2- + 8 H 3 O + . NO 3 - + 4 H 3 O + --> NO + 6 H 2 O Step 5: Balance charge by inserting e - (electrons) as a reactant or product in each half-reaction. Oxidation: CuS + 12 H 2 O ---> Cu 2+ + SO 4 2- + 8 H 3 O + + 8 e - . Reduction: NO 3 - + 4 H 3 O + + 3 e - --> NO + 6 H 2 O . Step 6: Multiply the two half-reactions by numbers chosen to make the number of electrons given off by the oxidation step equal to the number taken up by the reduction step. Then add the two half-reactions. If done correctly, the electrons should cancel out (equal numbers on the reactant and product sides of the overall reaction). If H 3 O + , H 2 O, or OH - appears on both sides of the final equation, cancel out the duplication also. Here the oxidation half-reaction must be multiplied by 3 (so that 24 electrons are produced) and the reduction half-reaction must by multiplied by 8 (so that the same 24 electrons are consumed). 3 CuS + 36 H 2 O ---> 3 Cu 2+ + 3 SO 4 2- + 24 H 3 O + + 24 e - 8 NO 3 - + 32 H 3 O + + 24 e - ---> 8 NO + 48 H 2 O Adding these two together gives the following equation: 3 CuS + 36 H 2 O + 8 NO 3 - + 8 H 3 O + ---> 3 Cu 2+ + 3 SO 4 2- + 8 NO + 48 H 2 O Step 7: Finally balancing both sides for excess of H 2 O (On each side -36) This gives you the following overall balanced equation at last: 3 CuS (s) + 8 NO 3 - (aq) + 8 H 3 O + (aq) ---> 3 Cu 2+ (aq) + 3 SO 4 2- (aq) + 8 NO (g) + 12 H 2 O (l)

6 0

3 years ago

Read 2 more answers

The elements least likely to form bonds are found in which group?

taurus [48]

Answer:

Noble gases are a unique set of elements in the periodic table because they don't naturally bond with other elements.

Explanation:

HAVE A GOOD DAY!

4 0

2 years ago