NVAEasyJEE Main 2025 · 24 January, Shift 1Entropy & Spontaneity

Chemistry Question from JEE Main 2025 · 24 January, Shift 1

Standard entropies of X2X_2, Y2Y_2 and XY5XY_5 are 70JK1mol170 \, \text{J} \, \text{K}^{-1} \, \text{mol}^{-1}, 50JK1mol150 \, \text{J} \, \text{K}^{-1} \, \text{mol}^{-1}, and 110JK1mol1110 \, \text{J} \, \text{K}^{-1} \, \text{mol}^{-1} respectively. The temperature in Kelvin at which the reaction

12X2+52Y2XY5ΔH=35kJ mol1\frac{1}{2} X_2 + \frac{5}{2} Y_2 \rightarrow XY_5 \quad \Delta H = -35 \, \text{kJ mol}^{-1}

will be at equilibrium is (nearest integer):

Answer & step-by-step solution

Sign in to reveal the correct answer, the full step-by-step solution, and the common mistakes for this question.

Practice more Entropy & Spontaneity questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step - free to start.

Related questions