{"id":84932,"date":"2025-06-01T02:56:24","date_gmt":"2025-06-01T02:56:24","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=84932"},"modified":"2025-06-01T02:56:24","modified_gmt":"2025-06-01T02:56:24","slug":"for-a-certain-reaction-delta-g0-45-kj-mol-and-delta-h0-9","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/for-a-certain-reaction-delta-g0-45-kj-mol-and-delta-h0-9\/","title":{"rendered":"For a certain reaction, $\\Delta G^0 = -45$ kJ\/mol and $\\Delta H^0 = -9"},"content":{"rendered":"

For a certain reaction, $\\Delta G^0 = -45$ kJ\/mol and $\\Delta H^0 = -90$ kJ\/mol at 0 \u00b0C. What is the minimum temperature at which the reaction will become spontaneous, assuming that $\\Delta H^0$ and $\\Delta S^0$ are independent of temperature?<\/p>\n

[amp_mcq option1=”273 K” option2=”298 K” option3=”546 K” option4=”596 K” correct=”option3″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC CDS-1 – 2019<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe spontaneity of a reaction is determined by the change in Gibbs Free Energy, $\\Delta G$. A reaction is spontaneous if $\\Delta G < 0$. The relationship between $\\Delta G$, $\\Delta H$, and $\\Delta S$ is given by $\\Delta G = \\Delta H - T\\Delta S$. We are given $\\Delta G^0 = -45$ kJ\/mol and $\\Delta H^0 = -90$ kJ\/mol at 0 \u00b0C (273 K). Assuming $\\Delta H^0$ and $\\Delta S^0$ are independent of temperature, we can calculate $\\Delta S^0$ at 273 K:\n$\\Delta G^0_{273} = \\Delta H^0 - (273 \\text{ K})\\Delta S^0$\n$-45 \\text{ kJ\/mol} = -90 \\text{ kJ\/mol} - 273\\Delta S^0$\n$273\\Delta S^0 = -90 + 45 = -45 \\text{ kJ\/mol}$\n$\\Delta S^0 = \\frac{-45}{273} \\text{ kJ\/(mol\u00b7K)}$\n\nThe reaction is spontaneous when $\\Delta G < 0$:\n$\\Delta H^0 - T\\Delta S^0 < 0$\n$-90 \\text{ kJ\/mol} - T \\left(\\frac{-45}{273} \\text{ kJ\/(mol\u00b7K)}\\right) < 0$\n$-90 + T \\left(\\frac{45}{273}\\right) < 0$\n$T \\left(\\frac{45}{273}\\right) < 90$\n$T < 90 \\times \\frac{273}{45}$\n$T < 2 \\times 273$\n$T < 546 \\text{ K}$\nSo, the reaction is spontaneous at temperatures below 546 K. The temperature at which $\\Delta G$ becomes zero (equilibrium) is $T = 546$ K. The question asks for the \"minimum temperature at which the reaction will become spontaneous\". While the phrasing is awkward for a reaction that is spontaneous below a certain temperature, it likely refers to the boundary temperature (546 K) where spontaneity starts or ends depending on the temperature direction, or potentially the lowest temperature among options where it is spontaneous. However, the calculation directly yields 546 K as the key temperature threshold.\n<\/section>\n
\n– Spontaneity requires $\\Delta G < 0$.\n- $\\Delta G = \\Delta H - T\\Delta S$.\n- Calculate $\\Delta S$ from the given data at 273 K.\n- Find the temperature range where $\\Delta G < 0$.\n- The transition temperature where $\\Delta G = 0$ is $T_{eq} = \\Delta H \/ \\Delta S$.\n<\/section>\n
\nFor reactions with $\\Delta H < 0$ and $\\Delta S < 0$, the reaction is spontaneous at low temperatures ($T < T_{eq}$) and non-spontaneous at high temperatures ($T > T_{eq}$). The boundary temperature is $T_{eq}$.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

For a certain reaction, $\\Delta G^0 = -45$ kJ\/mol and $\\Delta H^0 = -90$ kJ\/mol at 0 \u00b0C. What is the minimum temperature at which the reaction will become spontaneous, assuming that $\\Delta H^0$ and $\\Delta S^0$ are independent of temperature? [amp_mcq option1=”273 K” option2=”298 K” option3=”546 K” option4=”596 K” correct=”option3″] This question was previously … <\/p>\n

Detailed SolutionFor a certain reaction, $\\Delta G^0 = -45$ kJ\/mol and $\\Delta H^0 = -9<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1087],"tags":[1119,1096,1155],"class_list":["post-84932","post","type-post","status-publish","format-standard","hentry","category-upsc-cds-1","tag-1119","tag-chemistry","tag-physical-and-chemical-changes-solution","no-featured-image-padding"],"yoast_head":"\nFor a certain reaction, $\\Delta G^0 = -45$ kJ\/mol and $\\Delta H^0 = -9<\/title>\n<meta name=\"description\" content=\"The spontaneity of a reaction is determined by the change in Gibbs Free Energy, $Delta G$. A reaction is spontaneous if $Delta G < 0$. The relationship between $Delta G$, $Delta H$, and $Delta S$ is given by $Delta G = Delta H - TDelta S$. We are given $Delta G^0 = -45$ kJ\/mol and $Delta H^0 = -90$ kJ\/mol at 0 \u00b0C (273 K). Assuming $Delta H^0$ and $Delta S^0$ are independent of temperature, we can calculate $Delta S^0$ at 273 K: $Delta G^0_{273} = Delta H^0 - (273 text{ K})Delta S^0$ $-45 text{ kJ\/mol} = -90 text{ kJ\/mol} - 273Delta S^0$ $273Delta S^0 = -90 + 45 = -45 text{ kJ\/mol}$ $Delta S^0 = frac{-45}{273} text{ kJ\/(mol\u00b7K)}$ The reaction is spontaneous when $Delta G < 0$: $Delta H^0 - TDelta S^0 < 0$ $-90 text{ kJ\/mol} - T left(frac{-45}{273} text{ kJ\/(mol\u00b7K)}right) < 0$ $-90 + T left(frac{45}{273}right) < 0$ $T left(frac{45}{273}right) < 90$ $T < 90 times frac{273}{45}$ $T < 2 times 273$ $T < 546 text{ K}$ So, the reaction is spontaneous at temperatures below 546 K. The temperature at which $Delta G$ becomes zero (equilibrium) is $T = 546$ K. The question asks for the "minimum temperature at which the reaction will become spontaneous". While the phrasing is awkward for a reaction that is spontaneous below a certain temperature, it likely refers to the boundary temperature (546 K) where spontaneity starts or ends depending on the temperature direction, or potentially the lowest temperature among options where it is spontaneous. However, the calculation directly yields 546 K as the key temperature threshold. - Spontaneity requires $Delta G < 0$. - $Delta G = Delta H - TDelta S$. - Calculate $Delta S$ from the given data at 273 K. - Find the temperature range where $Delta G < 0$. - The transition temperature where $Delta G = 0$ is $T_{eq} = Delta H \/ Delta S$.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/for-a-certain-reaction-delta-g0-45-kj-mol-and-delta-h0-9\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"For a certain reaction, $\\Delta G^0 = -45$ kJ\/mol and $\\Delta H^0 = -9\" \/>\n<meta property=\"og:description\" content=\"The spontaneity of a reaction is determined by the change in Gibbs Free Energy, $Delta G$. A reaction is spontaneous if $Delta G < 0$. The relationship between $Delta G$, $Delta H$, and $Delta S$ is given by $Delta G = Delta H - TDelta S$. We are given $Delta G^0 = -45$ kJ\/mol and $Delta H^0 = -90$ kJ\/mol at 0 \u00b0C (273 K). Assuming $Delta H^0$ and $Delta S^0$ are independent of temperature, we can calculate $Delta S^0$ at 273 K: $Delta G^0_{273} = Delta H^0 - (273 text{ K})Delta S^0$ $-45 text{ kJ\/mol} = -90 text{ kJ\/mol} - 273Delta S^0$ $273Delta S^0 = -90 + 45 = -45 text{ kJ\/mol}$ $Delta S^0 = frac{-45}{273} text{ kJ\/(mol\u00b7K)}$ The reaction is spontaneous when $Delta G < 0$: $Delta H^0 - TDelta S^0 < 0$ $-90 text{ kJ\/mol} - T left(frac{-45}{273} text{ kJ\/(mol\u00b7K)}right) < 0$ $-90 + T left(frac{45}{273}right) < 0$ $T left(frac{45}{273}right) < 90$ $T < 90 times frac{273}{45}$ $T < 2 times 273$ $T < 546 text{ K}$ So, the reaction is spontaneous at temperatures below 546 K. The temperature at which $Delta G$ becomes zero (equilibrium) is $T = 546$ K. The question asks for the "minimum temperature at which the reaction will become spontaneous". While the phrasing is awkward for a reaction that is spontaneous below a certain temperature, it likely refers to the boundary temperature (546 K) where spontaneity starts or ends depending on the temperature direction, or potentially the lowest temperature among options where it is spontaneous. However, the calculation directly yields 546 K as the key temperature threshold. - Spontaneity requires $Delta G < 0$. - $Delta G = Delta H - TDelta S$. - Calculate $Delta S$ from the given data at 273 K. - Find the temperature range where $Delta G < 0$. - The transition temperature where $Delta G = 0$ is $T_{eq} = Delta H \/ Delta S$.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/for-a-certain-reaction-delta-g0-45-kj-mol-and-delta-h0-9\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T02:56:24+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"2 minutes\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"For a certain reaction, $\\Delta G^0 = -45$ kJ\/mol and $\\Delta H^0 = -9","description":"The spontaneity of a reaction is determined by the change in Gibbs Free Energy, $Delta G$. A reaction is spontaneous if $Delta G < 0$. The relationship between $Delta G$, $Delta H$, and $Delta S$ is given by $Delta G = Delta H - TDelta S$. We are given $Delta G^0 = -45$ kJ\/mol and $Delta H^0 = -90$ kJ\/mol at 0 \u00b0C (273 K). Assuming $Delta H^0$ and $Delta S^0$ are independent of temperature, we can calculate $Delta S^0$ at 273 K: $Delta G^0_{273} = Delta H^0 - (273 text{ K})Delta S^0$ $-45 text{ kJ\/mol} = -90 text{ kJ\/mol} - 273Delta S^0$ $273Delta S^0 = -90 + 45 = -45 text{ kJ\/mol}$ $Delta S^0 = frac{-45}{273} text{ kJ\/(mol\u00b7K)}$ The reaction is spontaneous when $Delta G < 0$: $Delta H^0 - TDelta S^0 < 0$ $-90 text{ kJ\/mol} - T left(frac{-45}{273} text{ kJ\/(mol\u00b7K)}right) < 0$ $-90 + T left(frac{45}{273}right) < 0$ $T left(frac{45}{273}right) < 90$ $T < 90 times frac{273}{45}$ $T < 2 times 273$ $T < 546 text{ K}$ So, the reaction is spontaneous at temperatures below 546 K. The temperature at which $Delta G$ becomes zero (equilibrium) is $T = 546$ K. The question asks for the "minimum temperature at which the reaction will become spontaneous". While the phrasing is awkward for a reaction that is spontaneous below a certain temperature, it likely refers to the boundary temperature (546 K) where spontaneity starts or ends depending on the temperature direction, or potentially the lowest temperature among options where it is spontaneous. However, the calculation directly yields 546 K as the key temperature threshold. - Spontaneity requires $Delta G < 0$. - $Delta G = Delta H - TDelta S$. - Calculate $Delta S$ from the given data at 273 K. - Find the temperature range where $Delta G < 0$. - The transition temperature where $Delta G = 0$ is $T_{eq} = Delta H \/ Delta S$.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/for-a-certain-reaction-delta-g0-45-kj-mol-and-delta-h0-9\/","og_locale":"en_US","og_type":"article","og_title":"For a certain reaction, $\\Delta G^0 = -45$ kJ\/mol and $\\Delta H^0 = -9","og_description":"The spontaneity of a reaction is determined by the change in Gibbs Free Energy, $Delta G$. A reaction is spontaneous if $Delta G < 0$. The relationship between $Delta G$, $Delta H$, and $Delta S$ is given by $Delta G = Delta H - TDelta S$. We are given $Delta G^0 = -45$ kJ\/mol and $Delta H^0 = -90$ kJ\/mol at 0 \u00b0C (273 K). Assuming $Delta H^0$ and $Delta S^0$ are independent of temperature, we can calculate $Delta S^0$ at 273 K: $Delta G^0_{273} = Delta H^0 - (273 text{ K})Delta S^0$ $-45 text{ kJ\/mol} = -90 text{ kJ\/mol} - 273Delta S^0$ $273Delta S^0 = -90 + 45 = -45 text{ kJ\/mol}$ $Delta S^0 = frac{-45}{273} text{ kJ\/(mol\u00b7K)}$ The reaction is spontaneous when $Delta G < 0$: $Delta H^0 - TDelta S^0 < 0$ $-90 text{ kJ\/mol} - T left(frac{-45}{273} text{ kJ\/(mol\u00b7K)}right) < 0$ $-90 + T left(frac{45}{273}right) < 0$ $T left(frac{45}{273}right) < 90$ $T < 90 times frac{273}{45}$ $T < 2 times 273$ $T < 546 text{ K}$ So, the reaction is spontaneous at temperatures below 546 K. The temperature at which $Delta G$ becomes zero (equilibrium) is $T = 546$ K. The question asks for the "minimum temperature at which the reaction will become spontaneous". While the phrasing is awkward for a reaction that is spontaneous below a certain temperature, it likely refers to the boundary temperature (546 K) where spontaneity starts or ends depending on the temperature direction, or potentially the lowest temperature among options where it is spontaneous. However, the calculation directly yields 546 K as the key temperature threshold. - Spontaneity requires $Delta G < 0$. - $Delta G = Delta H - TDelta S$. - Calculate $Delta S$ from the given data at 273 K. - Find the temperature range where $Delta G < 0$. - The transition temperature where $Delta G = 0$ is $T_{eq} = Delta H \/ Delta S$.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/for-a-certain-reaction-delta-g0-45-kj-mol-and-delta-h0-9\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T02:56:24+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"2 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/for-a-certain-reaction-delta-g0-45-kj-mol-and-delta-h0-9\/","url":"https:\/\/exam.pscnotes.com\/mcq\/for-a-certain-reaction-delta-g0-45-kj-mol-and-delta-h0-9\/","name":"For a certain reaction, $\\Delta G^0 = -45$ kJ\/mol and $\\Delta H^0 = -9","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T02:56:24+00:00","dateModified":"2025-06-01T02:56:24+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The spontaneity of a reaction is determined by the change in Gibbs Free Energy, $\\Delta G$. A reaction is spontaneous if $\\Delta G < 0$. The relationship between $\\Delta G$, $\\Delta H$, and $\\Delta S$ is given by $\\Delta G = \\Delta H - T\\Delta S$. We are given $\\Delta G^0 = -45$ kJ\/mol and $\\Delta H^0 = -90$ kJ\/mol at 0 \u00b0C (273 K). Assuming $\\Delta H^0$ and $\\Delta S^0$ are independent of temperature, we can calculate $\\Delta S^0$ at 273 K: $\\Delta G^0_{273} = \\Delta H^0 - (273 \\text{ K})\\Delta S^0$ $-45 \\text{ kJ\/mol} = -90 \\text{ kJ\/mol} - 273\\Delta S^0$ $273\\Delta S^0 = -90 + 45 = -45 \\text{ kJ\/mol}$ $\\Delta S^0 = \\frac{-45}{273} \\text{ kJ\/(mol\u00b7K)}$ The reaction is spontaneous when $\\Delta G < 0$: $\\Delta H^0 - T\\Delta S^0 < 0$ $-90 \\text{ kJ\/mol} - T \\left(\\frac{-45}{273} \\text{ kJ\/(mol\u00b7K)}\\right) < 0$ $-90 + T \\left(\\frac{45}{273}\\right) < 0$ $T \\left(\\frac{45}{273}\\right) < 90$ $T < 90 \\times \\frac{273}{45}$ $T < 2 \\times 273$ $T < 546 \\text{ K}$ So, the reaction is spontaneous at temperatures below 546 K. The temperature at which $\\Delta G$ becomes zero (equilibrium) is $T = 546$ K. The question asks for the "minimum temperature at which the reaction will become spontaneous". While the phrasing is awkward for a reaction that is spontaneous below a certain temperature, it likely refers to the boundary temperature (546 K) where spontaneity starts or ends depending on the temperature direction, or potentially the lowest temperature among options where it is spontaneous. However, the calculation directly yields 546 K as the key temperature threshold. - Spontaneity requires $\\Delta G < 0$. - $\\Delta G = \\Delta H - T\\Delta S$. - Calculate $\\Delta S$ from the given data at 273 K. - Find the temperature range where $\\Delta G < 0$. - The transition temperature where $\\Delta G = 0$ is $T_{eq} = \\Delta H \/ \\Delta S$.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/for-a-certain-reaction-delta-g0-45-kj-mol-and-delta-h0-9\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/for-a-certain-reaction-delta-g0-45-kj-mol-and-delta-h0-9\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/for-a-certain-reaction-delta-g0-45-kj-mol-and-delta-h0-9\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC CDS-1","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-cds-1\/"},{"@type":"ListItem","position":3,"name":"For a certain reaction, $\\Delta G^0 = -45$ kJ\/mol and $\\Delta H^0 = -9"}]},{"@type":"WebSite","@id":"https:\/\/exam.pscnotes.com\/mcq\/#website","url":"https:\/\/exam.pscnotes.com\/mcq\/","name":"MCQ and Quiz for Exams","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/exam.pscnotes.com\/mcq\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209","name":"rawan239","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","caption":"rawan239"},"sameAs":["https:\/\/exam.pscnotes.com"],"url":"https:\/\/exam.pscnotes.com\/mcq\/author\/rawan239\/"}]}},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/84932","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/comments?post=84932"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/84932\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=84932"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=84932"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=84932"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}