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Saurabh Bhai Chatterjee
Asked: 2022-11-11T04:45:07+05:30 In:

Greenhouse gas `CO_(2)` can be converted to `CO(g)` by the following reaction
`CO_(2)(g)+H_(2)(g) rarr CO_(2)+H_(2)O(g)` , termed as water gas reaction.
Calculate `DeltaG` for the reaction at 1000K `(DeltaH_(1000K)=35040 J “mol”^(-1) DeltaS_(1000K) =32.11 J “mol”^(-1)K^(1))`.

Greenhouse gas `CO_(2)` can be converted to `CO(g)` by the following reaction
`CO_(2)(g)+H_(2)(g) rarr CO_(2)+H_(2)O(g)` , termed as water gas reaction.
Calculate `DeltaG` for the reaction at 1000K `(DeltaH_(1000K)=35040 J “mol”^(-1) DeltaS_(1000K) =32.11 J “mol”^(-1)K^(1))`.
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  1. Correct Answer – `DeltaG=DeltaH-TDeltaS ” ” therefore DeltaG=35040- 1000 xx 32.11=2930 J`

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Jawahar Bajaj
Asked: 2022-10-29T16:58:52+05:30 In:

Greenhouse gas `CO_(2)` can be converted to `CO(g)` by the following reaction
`CO_(2)(g)+H_(2)(g) rarr CO_(2)+H_(2)O(g)` , termed as water gas reaction.
Calculate `DeltaH` at 1400 K using the given data for 1000K , assuming the `C_(p)^(@)` values remain constant in the given temoerature range.
`DeltaH=35040 J”mol”^(-1), C_(p)^(@)(CO_(2))=(42.31 + 10.09 xx 1^(-3)T) J “mol”^(-1)K_(1)`
`C_(P)^(@)(H_(2))=(27.40 + 3.20 xx 10^(-3)T) J”mol”^(-1)K_(1)`
`C_(P)^(@)(CO)=(28.34+ 4.14 xx 10^(-3)T)J”mol”^(-1)K^(-1)`
`C_(P)^(@)(H_(2)O)=(30.09 + 10.67 xx 10^(-3)T)J”mol^(-1)K^(-1)`

Greenhouse gas `CO_(2)` can be converted to `CO(g)` by the following reaction
`CO_(2)(g)+H_(2)(g) rarr CO_(2)+H_(2)O(g)` , termed as water gas reaction.
Calculate `DeltaH` at 1400 K using the given data for 1000K , assuming the `C_(p)^(@)` values remain constant in the given temoerature range.
`DeltaH=35040 J”mol”^(-1), C_(p)^(@)(CO_(2))=(42.31 + 10.09 xx 1^(-3)T) J “mol”^(-1)K_(1)`
`C_(P)^(@)(H_(2))=(27.40 + 3.20 xx 10^(-3)T) J”mol”^(-1)K_(1)`
`C_(P)^(@)(CO)=(28.34+ 4.14 xx 10^(-3)T)J”mol”^(-1)K^(-1)`
`C_(P)^(@)(H_(2)O)=(30.09 + 10.67 xx 10^(-3)T)J”mol^(-1)K^(-1)`
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  1. Correct Answer – `int_(H_(1))^(H_(2))dH=int_(T_(1))^(T_(2)) Cp.dT`
    `DeltaH_(2)-DeltaH_(1) = int_(T_(1))^(T_(2)) (28.35 + 4.14 xx 10^(-3)T + 30.09 + 10.67 xx 10^(-3)T)-(42.31 + 10.09 xx 10^(-3)T + 27.4 + 3.2xx 10^(-3)T)dT`
    `=int_(T_(1))^(T_(2)) (-11.28 + 1.52 xx 10^(-3)T)dT`
    `=-11.28 xx 400 + 1.52 xx ((1400^(2) – 1000^(2)))/(2) xx 10^(-3)`
    `DeltaH_(2) – 35040 =- 3782.4 ” ” therefore DeltaH_(2) = 31258 “Joule”`

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