/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q7E Describe the effect of each of t... [FREE SOLUTION] | 91影视

91影视

Describe the effect of each of the following on the rate of the reaction of magnesium metal with a solution of hydrochloric acid: the molarity of the hydrochloric acid, the temperature of the solution, and the size of the pieces of magnesium.

Short Answer

Expert verified

The effect on the rate of the reaction of magnesium metal with a solution of hydrochloric acid increases with the molarity of hydrochloric acid, increase with increased temperature and increase with the reactant size.

Step by step solution

01

Molarity of the hydrochloric acid

Molarity means the number of moles of solute present per litre

The rate of reaction is dependent on the frequency at which molecules collide. Increasing the molarity means increasing the concentration of molecules, then increasing the collision between the molecule and increasing the rate of reaction.

02

The temperature of the solution.

Temperature increases the number of particles to cross the activation energy barrier to starting making a product.

Rate of reaction increase with increased temperature because the molecule collides more frequently.

03

the size of the pieces of magnesium.

The rate of reaction depends on reactant size because the greater the surface area thus faster the rate of reaction.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

There are two molecules with the formula\({{\bf{C}}_{\bf{3}}}{{\bf{H}}_{\bf{6}}}\). Propene,\({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CH = C}}{{\bf{H}}_{\bf{2}}}\), is the monomer of the polymer polypropylene, which is used for indoor-outdoor carpets. Cyclopropane is used as an anaesthetic:

When heated to 499\({\bf{^\circ C}}\), cyclopropane rearranges (isomerizes) and forms propene with a rate constant of\({\bf{5}}{\bf{.95 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}{{\bf{s}}^{{\bf{ - 1}}}}\). What is the half-life of this reaction? What fraction of the cyclopropane remains after 0.75 h at 499.5\({\bf{^\circ C}}\)?

Use the PhET Reactions & Rates interactive simulation to simulate a system. On the 鈥淪ingle collision鈥 tab of the simulation applet, enable the 鈥淓nergy view鈥 by clicking the 鈥+鈥 icon. Select the first A + BC鉄禔B + C reaction (A is yellow, B is purple, and C is navy blue). Using the 鈥渟traight shot鈥 default option, try launching the A atom with varying amounts of energy. What changes when the Total Energy line at launch is below the transition state of the Potential Energy line? Why? What happens when it is above the transition state? Why?

Given the following reactions and the corresponding rate laws, in which of the reactions might the elementary reaction and the overall reaction be the same?\(\begin{aligned}{\rm{(a) C}}{{\rm{l}}_2}{\rm{ + CO }} \to {\rm{ C}}{{\rm{l}}_2}{\rm{CO}}\\{\rm{rate = }}k{{\rm{(C}}{{\rm{l}}_2}{\rm{)}}^{\frac{3}{2}}}{\rm{(CO)}}\\{\rm{(b) PC}}{{\rm{l}}_3}{\rm{ + C}}{{\rm{l}}_{\rm{2}}}{\rm{ }} \to {\rm{ PC}}{{\rm{l}}_{\rm{5}}}\\{\rm{rate = }}k{\rm{(PC}}{{\rm{l}}_{\rm{3}}}{\rm{) (C}}{{\rm{l}}_{\rm{2}}}{\rm{)}}\\{\rm{(c) 2NO + }}{{\rm{H}}_{\rm{2}}}{\rm{ }} \to {\rm{ }}{{\rm{N}}_{\rm{2}}}{\rm{ + }}{{\rm{H}}_{\rm{2}}}{\rm{O}}\\{\rm{rate = }}k{\rm{(NO)(}}{{\rm{H}}_{\rm{2}}}{\rm{)}}\\{\rm{(d) 2NO + }}{{\rm{O}}_{\rm{2}}}{\rm{ }} \to {\rm{ 2N}}{{\rm{O}}_{\rm{2}}}\\{\rm{rate = }}k{{\rm{(NO)}}^{\rm{2}}}{\rm{(}}{{\rm{O}}_{\rm{2}}}{\rm{)}}\\{\rm{(e) NO + }}{{\rm{O}}_{\rm{3}}}{\rm{ }} \to {\rm{ N}}{{\rm{O}}_{\rm{2}}}{\rm{ + }}{{\rm{O}}_{\rm{2}}}\\{\rm{rate = }}k{\rm{(NO)(}}{{\rm{O}}_{\rm{3}}}{\rm{)}}\end{aligned}\)

:How does an increase in temperature affect rate of reaction? Explain this effect in terms of the collision theory of the reaction rate

Regular flights of supersonic aircraft in the stratosphere are of concern because such aircraft produce nitric oxide, NO, as a by-product in the exhaust of their engines. Nitric oxide reacts with ozone, and it has been suggested that this could contribute to depletion of the ozone layer. The reaction \({\bf{NO + }}{{\bf{O}}_{\bf{3}}} \to {\bf{N}}{{\bf{O}}_{\bf{2}}}{\bf{ + }}{{\bf{O}}_{\bf{2}}}\) is first order with respect to both NO and \({{\bf{O}}_{\bf{3}}}\) with a rate constant of \({\bf{2}}{\bf{.20 \times 1}}{{\bf{0}}^{\bf{7}}}{\bf{mol}}{{\bf{L}}^{{\bf{ - 1}}}}{{\bf{s}}^{{\bf{ - 1}}}}\). What is the instantaneous rate of disappearance of NO when \(\left( {{\bf{NO}}} \right){\bf{ = 3}}{\bf{.3 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}{\bf{ M}}\) and \({\bf{(}}{{\bf{O}}_{\bf{3}}}{\bf{) = 5}}{\bf{.9 \times 1}}{{\bf{0}}^{{\bf{ - 7}}}}{\bf{ M}}\)?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.