Based on equation E=−2.178×10−18 J(n2Z2) certain conclusions are written. Which of them is not correct?
Held on 30 Apr 2013 · Verified 9 Jul 2026.
The negative sign in equation simply means that the energy of electron bound to the nucleus is lower than it would be if the electrons were at the infinite distance from the nucleus
Larger the value of n, the larger is the orbit radius
Equation can be used to calculate the change in energy when the electron changes orbit
For n=1 the electron has a more negative energy than it does for n=6 which means that the electron is more loosely bound in the smallest allowed orbit
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For the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = -198 kJ. Which condition favours forward reaction?
The unit of rate constant for a first-order reaction is:
For the reaction $\mathrm{A}(\mathrm{g}) \rightleftharpoons 2 \mathrm{~B}(\mathrm{~g})$, the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500 , at 1000 K . [Given : $\mathrm{R}=0.0831 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ ] $\mathrm{K}_{\mathrm{p}}$ for the reaction at 1000 K is
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$\begin{aligned} &\text { Consider the following compounds: }\\ &\mathrm{\underline{K}O}_2, \mathrm{H}_2 \mathrm{\underline{O}}_2 \text { and } \mathrm{H}_2 \mathrm{\underline{S}O}_4 \text {. } \end{aligned}$ The oxidation states of the underlined elements in them are, respectively,
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