Which of the following lines correctly show the temperature dependence of equilibrium constant K, for an exothermic reaction?

• A and D

• A and B

• B and C

• C and D

The combustion of benzene (l) gives CO₂(g) and H₂O(l). Given that heat of combustion of benzene at constant volume is –3263.9 kJ mol^–1 at 250C; the heat of combustion (in kJ mol^–1) of benzene at constant pressure will be (R = 8.314 JK^–1 mol^–1)

• –3267.6

• 4152.6

• –452.46

• 3260

Which of the following plots represent an exothermic reaction?

The condition for a reaction to occur spontaneously is

• $∆$H must be negative

• $∆$S must be negative

• $\left(∆\mathrm{H}-\mathrm{T}∆\mathrm{S}\right)$ must be negative

• $\left(∆\mathrm{H}+\mathrm{T}∆\mathrm{S}\right)$ must be negative

For the reaction, X₂Y₄ (l) → 2XY₂ (g) at 300 K, the values of $∆\mathrm{U}$ and $∆\mathrm{S}$ are 2 kcal and 20 cal K^-1 respectively. The value of $∆\mathrm{G}$ for the reaction is

• -3400 cal

• 3400 cal

• -2800 cal

• 2000 cal

The values of $∆$H and $∆$S of a certain reaction are -400 kJ mol^-1 and -20 kJ mol^-1K^-1 respectively. The temperature below which the reaction is spontaneous, is

• 100 K

• 20$°$C

• 20 K

• 120$°$C

337.

The enthalpy of vaporisation of a certain liquid at its boiling point of 35°C is 24.64 kJ mol^-1. The value of change in entropy for the process is

• 704 JK ^-1mol^-1

• 80 JK ^-1mol^-1

• 24.64 JK ^-1mol^-1

• 7.04 JK^-1mol^-1

Given that

$$\begin{gathered} \mathrm{C} + {\mathrm{O}}_2 \to {\mathrm{CO}}_2; ∆\mathrm{H}°= -\mathrm{x} \mathrm{kJ} \\ 2\mathrm{CO} + {\mathrm{O}}_2 \to 2{\mathrm{CO}}_2; ∆\mathrm{H}°= -\mathrm{y} \mathrm{kJ} \end{gathered}$$

The heat of formation of carbon monoxide will be

• $\frac{y- 2x}{2}$

• y + 2x

• 2x - y 

• $\frac{2\mathrm{x}-\mathrm{y}}{2}$

The value of $∆$H for cooling 2 mole of an ideal monoatomic gas from 225$°$C to 125$°$C at constant pressure will be [ given C_p= $\frac{5}{2}\mathrm{R}$].

• 250 R

• -500 R

• 500R

• -250 R

The change of entropy (dS) is defined as

• $\mathrm{dS} = \frac{\mathrm{\delta q}}{\mathrm{T}}$

• $\frac{\mathrm{dH}}{\mathrm{T}}$

• $\mathrm{dS} = \frac{{\mathrm{\delta q}}_{\mathrm{rev}}}{\mathrm{T}}$

• $\mathrm{dS} = \frac{\mathrm{dH} - \mathrm{dG}}{\mathrm{T}}$