/*! 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} Q54E Question: Hydrogen is prepared ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Question:Hydrogen is prepared commercially by the reaction of methane and water vapor at elevated temperatures. \(C{H_4}(g) + {H_2}O(g) \rightleftharpoons 3{H_2}(g) + CO(g)\)

Short Answer

Expert verified

The reaction of the elevated temperatures \(K = 6.28\)

Step by step solution

01

Define Hydrogen atom

A hydrogen atom is an atom of the chemical element hydrogen. The electrically neutral atom contains a single positively charged proton and a single negatively charged electron bound to the nucleus by the Coulomb force. Atomic hydrogen constitutes about 75% of the baryonic mass of the universe

02

The reaction of equilibrium constant reaction

We need to calculate equilibrium constant for following reaction:

\(C{H_4}(g) + {H_2}O(g) \to 3{H_2}(g) + CO(g)\)

this is a reverse reaction so it should be written with a double arrow. Equilibrium constant is calculated like this:

\(K = \frac{{c\left( {{H_2}} \right)_{eq}^3 \cdot c{{(CO)}_{eq}}}}{{c{{\left( {C{H_4}} \right)}_{eq}} \cdot c{{\left( {{H_2}O} \right)}_{eq}}}}\)

03

Solving the concentration of the compound \({C_{eq}}\) 

\(\begin{array}{*{20}{c}}{c{{\left( {{H_2}} \right)}_{eq}} = 1.15M}\\{c{{(CO)}_{eq}} = 0.126M}\\{c{{\left( {C{H_4}} \right)}_{eq}} = 0.126M}\\{c{{\left( {{H_2}O} \right)}_{eq}} = 0.242M}\end{array}\)

The constant equation\(\begin{array}{*{20}{c}}{K = \frac{{{{1.15}^3} \cdot 0.126}}{{0.126 \cdot 0.242}}}\\{K = 6.28}\end{array}\)

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

How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each?

a. \(2{\rm{N}}{{\rm{H}}_3}(g)\rightleftharpoons {{\rm{N}}_2}(g) + 3{{\rm{H}}_2}(g)\) \({\rm{\Delta }}H = 92{\rm{kJ}}\)

b. \({{\rm{N}}_2}(g) + {{\rm{O}}_2}(g)\rightleftharpoons 2{\rm{NO}}(g)\) \({\rm{\Delta }}H = 181{\rm{kJ}}\)

c. \(2{{\rm{O}}_3}(g)\rightleftharpoons 3{{\rm{O}}_2}(g)\) \({\rm{\Delta }}H = - 285{\rm{kJ}}\)

d.\({\rm{CaO(s) + C}}{{\rm{O}}_{\rm{2}}}{\rm{(g)}}\rightleftharpoons {\rm{CaC}}{{\rm{O}}_{\rm{3}}}{\rm{(s)}}\) \({\rm{\Delta }}H = - 176{\rm{kJ}}\)

At a temperature of 60 ̊C, the vapor pressure of water is 0.196atm. What is the value of the equilibrium constant Kp for the transformation at 60 ̊C? H2O (l)⇌ H2O(g)

Question: Calculate the pressures of NO, Cl2, and NOCl in an equilibrium mixture produced by the reaction of a starting mixture with 4.0 atm NO and 2.0 atm Cl2. (Hint: KP is small; assume the reverse reaction goes to completion then comes back to equilibrium.)

Which of the systems described in Exercise 13.15 give homogeneous equilibria? Which give heterogeneous equilibria?

(a) \(C{H_4}(g) + C{l_2}\rightleftharpoons C{H_3}CI(g) + HCI(g)\)

(b)\({N_2}(g) + {O_2}(g)\rightleftharpoons 2NO(g)\)

(c)\(2S{O_2}(\;g) + {O_2}(\;g)\rightleftharpoons 2S{O_3}(\;g)\)

(d)\(BaS{O_3}(s)\rightleftharpoons BaO(s) + S{O_2}(g)\)

(e) \({P_4}(g) + 5{O_2}(g)\rightleftharpoons{P_4}{O_{10}}(s)\)

(f)\(B{r_2}(\;g)\rightleftharpoons 2Br(g)\)

(g) \(C{H_4}(g) + 2{O_2}(g)\rightleftharpoons C{O_2}(g) + 2{H_2}O(l)\)

(h) \(CuS{O_4} \times 5{H_2}O(s)\rightleftharpoons CuS{O_4}(s) + 5{H_2}O(g)\)

Question:Calculate the value of the equilibrium constant \({K_P}\) for the reaction \(2NO(g) + C{l_2}(g) \rightleftharpoons 2NOCl(g)\) from these equilibrium pressures: NO, \(0.050atm;C{l_2},0.30atm;NOCl,1.2atm\)

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.