Given below are two reactions, involved in the commercial production of dihydrogen (H2). The two reactions are carried out at temperature “T1 ” and “T2 ” respectively
\(C(S)+H_2O(g)\underrightarrow {T_1}\;CO(g)+H_2(g)\)
\(CO(g)+H_2O(g)\underrightarrow{T_2}\;CO_2(g)+H_2(g)\)
The temperature \(T_1\) and \(T_2\) are correctly related as
\(C(S)+H_2O(g)\underrightarrow {T_1}\;CO(g)+H_2(g)\)
\(CO(g)+H_2O(g)\underrightarrow{T_2}\;CO_2(g)+H_2(g)\)
The temperature \(T_1\) and \(T_2\) are correctly related as
Show Hint
The water-gas reaction needs very high temperatures. The water-gas shift reac tion, aided by a catalyst, can operate effectively at lower temperatures.
- \(T_1>T_2\)
- \(T_1<T_2\)
- T1=T2
- T1=100 K, T2=1270 K
The Correct Option is A
Solution and Explanation
Water-Gas Reaction
The water-gas reaction requires high temperatures for the reaction to proceed effectively:
Reaction Temperature: \( T_1 \approx 1270 \, \text{K} \)
Water-Gas Shift Reaction
The water-gas shift reaction is typically carried out at lower temperatures in the presence of a catalyst:
Reaction Temperature: \( T_2 \approx 673 \, \text{K} \)
Conclusion:
The temperature for the water-gas reaction (\( T_1 \)) is significantly higher than the temperature for the water-gas shift reaction (\( T_2 \)). Therefore:
\( T_1 > T_2 \)
Top Questions on Law Of Chemical Equilibrium And Equilibrium Constant
- For the reaction $ N_2 + 3H_2 \rightleftharpoons 2NH_3 $, if initially 1 mole of $ N_2 $ and 3 moles of $ H_2 $ are taken and at equilibrium 0.4 moles of $ NH_3 $ are formed, find the equilibrium concentration of $ H_2 $.
- BITSAT - 2025
- Chemistry
- Law Of Chemical Equilibrium And Equilibrium Constant
- For the reaction \( \mathrm{N_2} + 3\mathrm{H_2} \rightleftharpoons 2\mathrm{NH_3} \), the equilibrium constant \(K_c\) is 0.5 at a certain temperature. If initially, 1 mole of \(\mathrm{N_2}\) and 3 moles of \(\mathrm{H_2}\) are placed in a 1 L container, what is the equilibrium concentration of \(\mathrm{NH_3}\)?
- BITSAT - 2025
- Chemistry
- Law Of Chemical Equilibrium And Equilibrium Constant
The equilibrium constant for decomposition of $ H_2O $ (g) $ H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2} O_2(g) \quad (\Delta G^\circ = 92.34 \, \text{kJ mol}^{-1}) $ is $ 8.0 \times 10^{-3} $ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation ($ \alpha $) of water is _____ $\times 10^{-2}$ (nearest integer value). [Assume $ \alpha $ is negligible with respect to 1]
- JEE Main - 2025
- Chemistry
- Law Of Chemical Equilibrium And Equilibrium Constant
- Given below are two statements:
Statement I: A catalyst cannot alter the equilibrium constant ($ K_c $) of the reaction, temperature remaining constant.
Statement II: A homogeneous catalyst can change the equilibrium composition of a system, temperature remaining constant.
In the light of the above statements, choose the correct answer from the options given below.- JEE Main - 2025
- Chemistry
- Law Of Chemical Equilibrium And Equilibrium Constant
- In the following system, $ PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) $ at equilibrium, upon addition of xenon gas at constant T and p, the concentration of
- JEE Main - 2025
- Chemistry
- Law Of Chemical Equilibrium And Equilibrium Constant
Questions Asked in JEE Main exam
- If the equation of the parabola with vertex $\left(\frac{3}{2},\, 3\right)$ and the directrix $x + 2y = 0$ is
$$ ax^2 + by^2 - cxy - 30x - 60y + 225 = 0, $$ then $\alpha + \beta + \gamma$ is equal to:- JEE Main - 2025
- Calculus
- In a first order decomposition reaction, the time taken for the decomposition of reactant to one fourth and one eighth of its initial concentration are $ t_1 $ and $ t_2 $ (s), respectively. The ratio $ t_1 / t_2 $ will be:
- JEE Main - 2025
- Chemical Kinetics
- HA $ (aq) \rightleftharpoons H^+ (aq) + A^- (aq) $ The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is 0.20 °C. The dissociation constant for the acid is Given: $ K_f(H_2O) = 1.8 \, \text{K kg mol}^{-1} $, molality ≡ molarity
- JEE Main - 2025
- Colligative Properties
20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is _____ M. (Nearest Integer value) (Given : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol$^{-1}$)
- JEE Main - 2025
- Stoichiometry and Stoichiometric Calculations
- Which of the following binary mixture does not show the behavior of minimum boiling azeotropes?
- JEE Main - 2025
- Solutions