Changes in pressure for an equilibrium reaction containing a solid/liquid on one side only
For a reaction where there is only one solid/liquid present, but gases present as both reactants and products (are there any examples of this type of reaction?):
$$ce{A(s) + B(g) <=> C(g)},$$
if pressure is increased, it will favour the formation of the side not containing the solid/liquid (e.g. the forward reaction favoured), so as to decrease pressure of system – since one mole of a gas molecules has a considerably lower volume than a mole of any solid or liquid (far more spread out).
And the opposite if pressure decreased.
E.g. all that is considered is which side the solid/liquid is on, not the moles of gas present on either side. Is this true?
physical-chemistry equilibrium pressure
edited 3 hours ago
andselisk
13.2k64698
asked 3 hours ago
frog0101
111
New contributor
frog0101 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
For a reaction where there is only one solid/liquid present, but gases present as both reactants and products (are there any examples of this type of reaction?):
$$ce{A(s) + B(g) <=> C(g)},$$
if pressure is increased, it will favour the formation of the side not containing the solid/liquid (e.g. the forward reaction favoured), so as to decrease pressure of system – since one mole of a gas molecules has a considerably lower volume than a mole of any solid or liquid (far more spread out).
And the opposite if pressure decreased.
E.g. all that is considered is which side the solid/liquid is on, not the moles of gas present on either side. Is this true?
physical-chemistry equilibrium pressure
edited 3 hours ago
andselisk
13.2k64698
asked 3 hours ago
frog0101
111
New contributor
frog0101 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
For a reaction where there is only one solid/liquid present, but gases present as both reactants and products (are there any examples of this type of reaction?):
$$ce{A(s) + B(g) <=> C(g)},$$
if pressure is increased, it will favour the formation of the side not containing the solid/liquid (e.g. the forward reaction favoured), so as to decrease pressure of system – since one mole of a gas molecules has a considerably lower volume than a mole of any solid or liquid (far more spread out).
And the opposite if pressure decreased.
E.g. all that is considered is which side the solid/liquid is on, not the moles of gas present on either side. Is this true?
physical-chemistry equilibrium pressure
edited 3 hours ago
andselisk
13.2k64698
asked 3 hours ago
frog0101
111
New contributor
frog0101 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
For a reaction where there is only one solid/liquid present, but gases present as both reactants and products (are there any examples of this type of reaction?):
$$ce{A(s) + B(g) <=> C(g)},$$
if pressure is increased, it will favour the formation of the side not containing the solid/liquid (e.g. the forward reaction favoured), so as to decrease pressure of system – since one mole of a gas molecules has a considerably lower volume than a mole of any solid or liquid (far more spread out).
And the opposite if pressure decreased.
E.g. all that is considered is which side the solid/liquid is on, not the moles of gas present on either side. Is this true?
physical-chemistry equilibrium pressure
physical-chemistry equilibrium pressure
edited 3 hours ago
andselisk
13.2k64698
asked 3 hours ago
frog0101
111
New contributor
frog0101 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
andselisk
13.2k64698
asked 3 hours ago
frog0101
111
New contributor
frog0101 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
andselisk
13.2k64698
edited 3 hours ago
andselisk
13.2k64698
edited 3 hours ago
andselisk
13.2k64698
13.2k64698
asked 3 hours ago
frog0101
111
New contributor
frog0101 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
frog0101
111
asked 3 hours ago
frog0101
111
111
New contributor
frog0101 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
frog0101 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
frog0101 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
There are many examples of such reactions, for instance:
$$
begin{align}
ce{S (s) + O2 (g) &<=> SO2 (g)}\
ce{C (s) + 2 H2 (g) &<=> CH4 (g)}\
ce{Ni (s) + 4 CO (g) &<=> Ni(CO)4 (g)}
end{align}
$$
However, your assumption
all that is considered is which side the solid/liquid is on, not the moles of gas present on either side
is not correct. Since shift of equilibrium according to Le Chatelier’s principle depends on partial pressures, which in turn, rely on molar ratio of gaseous reactants and products, decrease/increase of pressure doesn't depend on which side of the reaction solid/liquid participant is located; instead, it is defined for each system individually based on ratio between gaseous products and reactants.
answered 3 hours ago
andselisk
13.2k64698
add a comment |
Well the reaction will favor the side not containing the solid/liquid, this is not true. It will favor the side where number of gaseous moles is less. This will be found giving a balanced chemical reaction.
answered 3 hours ago
harshit54
1108
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "431"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
frog0101 is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f107381%2fchanges-in-pressure-for-an-equilibrium-reaction-containing-a-solid-liquid-on-one%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
There are many examples of such reactions, for instance:
$$
begin{align}
ce{S (s) + O2 (g) &<=> SO2 (g)}\
ce{C (s) + 2 H2 (g) &<=> CH4 (g)}\
ce{Ni (s) + 4 CO (g) &<=> Ni(CO)4 (g)}
end{align}
$$
However, your assumption
all that is considered is which side the solid/liquid is on, not the moles of gas present on either side
is not correct. Since shift of equilibrium according to Le Chatelier’s principle depends on partial pressures, which in turn, rely on molar ratio of gaseous reactants and products, decrease/increase of pressure doesn't depend on which side of the reaction solid/liquid participant is located; instead, it is defined for each system individually based on ratio between gaseous products and reactants.
answered 3 hours ago
andselisk
13.2k64698
add a comment |
There are many examples of such reactions, for instance:
$$
begin{align}
ce{S (s) + O2 (g) &<=> SO2 (g)}\
ce{C (s) + 2 H2 (g) &<=> CH4 (g)}\
ce{Ni (s) + 4 CO (g) &<=> Ni(CO)4 (g)}
end{align}
$$
However, your assumption
all that is considered is which side the solid/liquid is on, not the moles of gas present on either side
is not correct. Since shift of equilibrium according to Le Chatelier’s principle depends on partial pressures, which in turn, rely on molar ratio of gaseous reactants and products, decrease/increase of pressure doesn't depend on which side of the reaction solid/liquid participant is located; instead, it is defined for each system individually based on ratio between gaseous products and reactants.
answered 3 hours ago
andselisk
13.2k64698
add a comment |
There are many examples of such reactions, for instance:
$$
begin{align}
ce{S (s) + O2 (g) &<=> SO2 (g)}\
ce{C (s) + 2 H2 (g) &<=> CH4 (g)}\
ce{Ni (s) + 4 CO (g) &<=> Ni(CO)4 (g)}
end{align}
$$
However, your assumption
all that is considered is which side the solid/liquid is on, not the moles of gas present on either side
is not correct. Since shift of equilibrium according to Le Chatelier’s principle depends on partial pressures, which in turn, rely on molar ratio of gaseous reactants and products, decrease/increase of pressure doesn't depend on which side of the reaction solid/liquid participant is located; instead, it is defined for each system individually based on ratio between gaseous products and reactants.
answered 3 hours ago
andselisk
13.2k64698
There are many examples of such reactions, for instance:
$$
begin{align}
ce{S (s) + O2 (g) &<=> SO2 (g)}\
ce{C (s) + 2 H2 (g) &<=> CH4 (g)}\
ce{Ni (s) + 4 CO (g) &<=> Ni(CO)4 (g)}
end{align}
$$
However, your assumption
all that is considered is which side the solid/liquid is on, not the moles of gas present on either side
is not correct. Since shift of equilibrium according to Le Chatelier’s principle depends on partial pressures, which in turn, rely on molar ratio of gaseous reactants and products, decrease/increase of pressure doesn't depend on which side of the reaction solid/liquid participant is located; instead, it is defined for each system individually based on ratio between gaseous products and reactants.
answered 3 hours ago
andselisk
13.2k64698
edited 2 hours ago
answered 3 hours ago
andselisk
13.2k64698
answered 3 hours ago
andselisk
13.2k64698
answered 3 hours ago
andselisk
13.2k64698
13.2k64698
add a comment |
add a comment |
Well the reaction will favor the side not containing the solid/liquid, this is not true. It will favor the side where number of gaseous moles is less. This will be found giving a balanced chemical reaction.
answered 3 hours ago
harshit54
1108
add a comment |
Well the reaction will favor the side not containing the solid/liquid, this is not true. It will favor the side where number of gaseous moles is less. This will be found giving a balanced chemical reaction.
answered 3 hours ago
harshit54
1108
add a comment |
Well the reaction will favor the side not containing the solid/liquid, this is not true. It will favor the side where number of gaseous moles is less. This will be found giving a balanced chemical reaction.
answered 3 hours ago
harshit54
1108
Well the reaction will favor the side not containing the solid/liquid, this is not true. It will favor the side where number of gaseous moles is less. This will be found giving a balanced chemical reaction.
answered 3 hours ago
harshit54
1108
answered 3 hours ago
harshit54
1108
answered 3 hours ago
harshit54
1108
answered 3 hours ago
harshit54
1108
1108
add a comment |
add a comment |
frog0101 is a new contributor. Be nice, and check out our Code of Conduct.
frog0101 is a new contributor. Be nice, and check out our Code of Conduct.
frog0101 is a new contributor. Be nice, and check out our Code of Conduct.
frog0101 is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Chemistry Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f107381%2fchanges-in-pressure-for-an-equilibrium-reaction-containing-a-solid-liquid-on-one%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
