/*! 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} Q70E Carbon reacts with water vapor a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Carbon reacts with water vapor at elevated temperatures

\(C(s) + {H_2}O(g) \rightleftharpoons CO(g) + {H_2}(g)\)

\({K_c} = 0.2at100{0^o}C\)

What is the concentration of CO in an equilibrium mixture with \(\left[ {{H_2}O} \right] = 0.500M\)at \(100{0^0}C\)?

Short Answer

Expert verified

The equilibrium concentration of CO is 0.316M.

Step by step solution

01

Find the equilibrium concentration of CO:

Let us assume the concentration of CO and \({H_2}\)is X

02

Find out the value of X

\(\begin{array}{c}{K_C} = \frac{{[{\rm{CO}}] \times \left[ {{{\rm{H}}_2}} \right]}}{{\left[ {{{\rm{H}}_2}{\rm{O}}} \right]}}\\{K_C} = \frac{{X \times X}}{{\left[ {{{\rm{H}}_2}{\rm{O}}} \right]}}\\{X^2} = {K_C} \times \left[ {{{\rm{H}}_2}{\rm{O}}} \right]\\{X^2} = 0.2 \times 0.5\end{array}\)

\(\begin{array}{c}{X^2} = 0.1\\X = 0.316\,M\\\left[ {CO} \right] = 0.316\,M\end{array}\)

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 91Ó°ÊÓ!

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

A necessary step in the manufacture of sulfuric acid is the formation of sulfur trioxide (\({\rm{S}}{{\rm{O}}_3}\)), from sulfur dioxide (\({\rm{S}}{{\rm{O}}_2}\)), and oxygen (\({{\rm{O}}_2}\)), shown here.

\(2{\text{S}}{{\text{O}}_2}(g) + {{\text{O}}_2}(g) \rightleftharpoons 2{\text{S}}{{\text{O}}_3}(g)\)

At high temperatures, the rate of formation of \({\rm{S}}{{\rm{O}}_3}\)is higher, but the equilibrium amount (concentration or partial pressure) of \({\rm{S}}{{\rm{O}}_3}\) is lower than it would be at lower temperatures.

(a) Does the equilibrium constant for the reaction increase, decrease, or remain about the same as the temperature increases?

(b) Is the reaction endothermic or exothermic?

Calculate the number of moles of \(HI\) that are at equilibrium with \(1.25mol\)of \({H_2}\)and \(1.25\,mol\)of \({I_2}\) in a \(5.00\,L\)flask at \(44{8^0}C\).\({H_2} + {I_2} \rightleftharpoons 2HI\)

\({K_c} = 50.2\,at\,44{8^o}C\)

Cobalt metal can be prepared by reducing cobalt (II) oxide with carbon monoxide.

\(CoO(s) + CO(g) \rightleftharpoons Co(s) + C{O_2}(g)\)

\({K_c} = 4.90 \times 1{0^2}at55{0^o}C\)

What concentration of \(CO\) remains in an equilibrium mixture with \(\left[ {C{O_2}} \right] = 0.100M\)

Why are there no changes specified for Ni What property of Ni does change?

Question:What is the value of the equilibrium constant at \(50{0^o}C\) for the formation of \(N{H_3}\)according to the following equation? N2(g) + 3H2(g) ⇌ 2NH3(g)

An equilibrium mixture of \(N{H_3}(g)\) \({H_2}(g)\) and \({N_2}(g)\) at \(50{0^o}C\) was found to contain\(1.35M{H_2},1.15M{N_2}\)and \(4.12\)\( \times 1{0^{ - 1}}MN{H_3}\)

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.