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)`

Answer

To calculate the change in enthalpy (ΔH) at 1400 K for the reaction:

\[ \text{CO}_2(g) + \text{H}_2(g) \rightarrow \text{CO}(g) + \text{H}_2\text{O}(g) \]

we will use the given data and the assumption that the heat capacities (C_p) remain constant over the temperature range.

### Step-by-Step Solution:

1. **Identify the given data:**
- ΔH at 1000 K = 35040 J/mol
- C_p for CO₂ = \( 42.31 + 10.09 \times 10^{-3}T \) J/mol·K
- C_p for H₂ = \( 27.40 + 3.20 \times 10^{-3}T \) J/mol·K
- C_p for CO = \( 28.34 + 4.14 \times 10^{-3}T \) J/mol·K
- C_p for H₂O = \( 30.09 + 10.67 \times 10^{-3}T \) J/mol·K

2. **Calculate C_p for reactants and products at 1400 K:**
- For CO₂:
\[ C_{p}(\text{CO}_2) = 42.31 + 10.09 \times 10^{-3} \times 1400 = 42.31 + 14.126 = 56.436 \text{ J/mol·K} \]
- For H₂:
\[ C_{p}(\text{H}_2) = 27.40 + 3.20 \times 10^{-3} \times 1400 = 27.40 + 4.48 = 31.88 \text{ J/mol·K} \]
- For CO:
\[ C_{p}(\text{CO}) = 28.34 + 4.14 \times 10^{-3} \times 1400 = 28.34 + 5.796 = 34.136 \text{ J/mol·K} \]
- For H₂O:
\[ C_{p}(\text{H}_2O) = 30.09 + 10.67 \times 10^{-3} \times 1400 = 30.09 + 14.938 = 45.028 \text{ J/mol·K} \]

3. **Calculate ΔC_p:**
\[ ΔC_p = [C_p(\text{CO}) + C_p(\text{H}_2O)] - [C_p(\text{CO}_2) + C_p(\text{H}_2)] \]
\[ ΔC_p = [34.136 + 45.028] - [56.436 + 31.88] \]
\[ ΔC_p = 79.164 - 88.316 = -9.152 \text{ J/mol·K} \]

4. **Calculate ΔH at 1400 K using the formula:**
\[ ΔH(T_2) = ΔH(T_1) + ΔC_p \cdot (T_2 - T_1) \]
Where \( T_1 = 1000 \text{ K} \) and \( T_2 = 1400 \text{ K} \):
\[ ΔH(1400) = 35040 + (-9.152) \cdot (1400 - 1000) \]
\[ ΔH(1400) = 35040 - 9.152 \cdot 400 \]
\[ ΔH(1400) = 35040 - 3660.8 = 31379.2 \text{ J/mol} \]

5. **Final Result:**
\[ ΔH(1400) \approx 31379.2 \text{ J/mol} \]