The enthalpy change for the reaction of $50.00 \ mL$ of ethylene with $50.00 \ mL$ of $H_2$ at $1.5 \ atm$ pressure is $\Delta H = -0.31 \ kJ$. The value of $\Delta E$ will be (in $kJ$)

The $\Delta H$ for vaporisation of a liquid is $20 \, kJ/mol$. Assuming ideal behaviour,the change in internal energy for the vaporisation of $1 \, mole$ of the liquid at $60^{\circ} C$ and $1 \, bar$ is close to $.... \, kJ/mol$

For a certain thermochemical reaction $M \rightarrow N$ at $T = 400 \ K$,$\Delta H^{\ominus} = 77.2 \ kJ \ mol^{-1}$ and $\Delta S = 122 \ J \ K^{-1} \ mol^{-1}$,the value of $\log K$ is $ . . . . . . \times 10^{-1}$.

One mole of an ideal gas at $900 \ K$ undergoes two reversible processes,$I$ followed by $II$,as shown in the graph. If the work done by the gas in the two processes is the same,the value of $\ln \frac{V_3}{V_2}$ is. . . . . . . . ($U$: internal energy,$S$: entropy,$p$: pressure,$V$: volume,$R$: gas constant). (Given: molar heat capacity at constant volume,$C_{V, m}$ of the gas is $\frac{5}{2} R$)

The enthalpy of combustion of carbon monoxide at $17 ^\circ C$ and constant volume is $-283.3 \ kJ$. Calculate the enthalpy of combustion at constant pressure in $kJ$.

The heat associated with the combustion of liquid benzene at constant volume is $-3268 \ kJ \ mol^{-1}$. Calculate the change in enthalpy when this reaction occurs at $300 \ K$ $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$.