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If enthalpy of overall reaction `XrarrY` along one route is `Delta_(r)H and Delta_(r)H_(1),Delta_(r)H_(2),Delta_(r)H_(3)`…. Representing enthalpies of reactions leading to same product Y then `Delta_(r)H` is

A

(a) `Delta_(r)H=Delta_(r)H_(1)+Delta_(r)H_(2)+Delta_(r)H_(3)…`

B

(b) `Delta_(r)H=Delta_(r)H_(1)xxDelta_(r)H_(2)xxDelta_(r)H_(3)…`

C

(c) `Delta_(r)H=Delta_(r)H_(1)+Delta_(r)H_(2)-Delta_(r)H_(3)…`

D

(d) `Delta_(r)H=(Delta_(r)H_(1)xxDelta_(r)H_(2)xxDelta_(r)H_(3))/(2)…`

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The correct Answer is:
To solve the problem, we need to establish the relationship between the enthalpy change of an overall reaction and the enthalpy changes of its individual steps. The question states that the enthalpy change for the overall reaction \( X \rightarrow Y \) is \( \Delta_r H \), and it can also be expressed as the sum of the enthalpy changes of multiple steps \( \Delta H_1, \Delta H_2, \Delta H_3, \ldots \). ### Step-by-Step Solution: 1. **Understanding the Reaction**: - We have a reaction where substance \( X \) is converted to substance \( Y \). - The enthalpy change for this overall reaction is denoted as \( \Delta_r H \). 2. **Identifying the Multi-Step Pathway**: - The reaction can occur through multiple intermediate steps. For example: - \( X \rightarrow P \) (with enthalpy change \( \Delta H_1 \)) - \( P \rightarrow Q \) (with enthalpy change \( \Delta H_2 \)) - \( Q \rightarrow Y \) (with enthalpy change \( \Delta H_3 \)) 3. **Applying Hess's Law**: - According to Hess's Law of Constant Heat Summation, the total enthalpy change for a reaction is the same regardless of the pathway taken. - This means that the enthalpy change for the overall reaction \( \Delta_r H \) can be expressed as the sum of the enthalpy changes of the individual steps: \[ \Delta_r H = \Delta H_1 + \Delta H_2 + \Delta H_3 + \ldots \] 4. **Conclusion**: - Therefore, the relationship between the overall enthalpy change \( \Delta_r H \) and the enthalpy changes of the individual steps \( \Delta H_1, \Delta H_2, \Delta H_3 \) is given by: \[ \Delta_r H = \sum_{i=1}^{n} \Delta H_i \] - This confirms that the enthalpy change for the overall reaction is equal to the sum of the enthalpy changes for each step leading to the same product. ### Final Answer: \[ \Delta_r H = \Delta H_1 + \Delta H_2 + \Delta H_3 + \ldots \]
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Standard Gibb's energy of reaction (Delta_(r )G^(@)) at a certain temperature can be computed Delta_(r )G^(@)=Delta_(r)H^(@)-T.Delta_(r )S^(@) and the change in the value of Delta_(r)H^(@) and Delta_(r)S^(@) for a reaction with temperature can be computed as follows : Delta_(r )H_(T_(2))^(@)-Delta_(r )H_(T_(1))^(@)=Delta_(r )C_(p)^(@)(T_(2)-T_(1)) Delta_(r )S_(T_(2))^(@)-Delta_(r )S_(T_(1))^(@)=Delta_(r )C_(p)^(@)ln.(T_(2)/T_(1)) " "Delta_(r )G^(@)=Delta_(r)H^(@)-T.Delta_(r)S^(@) and " by "Delta_(r )G^(@)=-"RT " ln K_(eq) . Consider the following reaction : CO(g)+2H_(2)(g)iffCH_(3)OH(g) Given : Delta_(f)H^(@)(CH_(3)OH,g)=-201 " kJ"//"mol", " "Delta_(f)H^(@)(CO,g)=-114" kJ"//"mol" S^(@)(CH_(3)OH,g)=240" J"//"K-mol, "S^(@)(H_(2),g)=29" JK"^(-1)" mol"^(-1) S^(@)(CO,g)=198 " J"//"mol-K, "C_(p,m)^(@)(H_(2))=28.8 " J"//"mol-K" C_(p,m)^(@)(CO)=29.4 " J"//"mol-K, "C_(p,m)^(@)(CH_(3)OH)=44 " J"//"mol-K" and " "ln ((320)/(300))=0.06 , all data at 300 K Delta_(r )S^(@) at 300 K for the reaction is :

Standard Gibb's energy of reaction (Delta_(r )G^(@)) at a certain temperature can be computed Delta_(r )G^(@)=Delta_(r)H^(@)-T.Delta_(r )S^(@) and the change in the value of Delta_(r)H^(@) and Delta_(r)S^(@) for a reaction with temperature can be computed as follows : Delta_(r )H_(T_(2))^(@)-Delta_(r )H_(T_(1))^(@)=Delta_(r )C_(p)^(@)(T_(2)-T_(1)) Delta_(r )S_(T_(2))^(@)-Delta_(r )S_(T_(1))^(@)=Delta_(r )C_(p)^(@)ln.(T_(2)/T_(1)) " "Delta_(r )G^(@)=Delta_(r)H^(@)-T.Delta_(r)S^(@) and " by "Delta_(r )G^(@)=-"RT " ln K_(eq) . Consider the following reaction : CO(g)+2H_(2)(g)iffCH_(3)OH(g) Given : Delta_(f)H^(@)(CH_(3)OH,g)=-201 " kJ"//"mol", " "Delta_(f)H^(@)(CO,g)=-114" kJ"//"mol" S^(@)(CH_(3)OH,g)=240" J"//"K-mol, "S^(@)(H_(2),g)=29" JK"^(-1)" mol"^(-1) S^(@)(CO,g)=198 " J"//"mol-K, "C_(p,m)^(@)(H_(2))=28.8 " J"//"mol-K" C_(p,m)^(@)(CO)=29.4 " J"//"mol-K, "C_(p,m)^(@)(CH_(3)OH)=44 " J"//"mol-K" and " "ln ((320)/(300))=0.06 , all data at 300 K Delta_(r )H^(@) at 320 K is :

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A: Total enthalpy change of a multistep process is sum of Delta H_(1) + Delta H_(2) + Delta H_(3) + .... R: When heat is absorbed by the system, the sign of q is taken to be negative.

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