When 400 mL of \(0.2 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) solution is mixed with 600 mL of of 0.1 M NaOH solution the increase in temperature of the final solution is \_\_\_\_ \(\times 10^{-2} \mathrm{~K}\). (Round off to the Nearest Integer). [Use: \(\mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{H}_{2} \mathrm{O}: \Delta_{\gamma} \mathrm{H}=-57.1 \mathrm{kJmol}^{-1}\) Specific heat of \(\mathrm{H}_{2} \mathrm{O}=4.18 \mathrm{JK}^{-1} \mathrm{~g}^{-1}\) Density of \(\mathrm{H}_{2} \mathrm{O}=1.0 \mathrm{gcm}^{-3}\) Assume no change in volume of solution on mixing]

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Chemical Thermodynamics

When 400 mL of

0.2 M H_{2} SO_{4}

solution is mixed with 600 mL of of 0.1 M NaOH solution the increase in temperature of the final solution is ____

\times 1 0^{- 2} K

.
(Round off to the Nearest Integer).
[Use:

H^{+} (aq) + OH^{-} (aq) \to H_{2} O : Δ_{γ} H = - 57.1 kJmol^{- 1}

Specific heat of

H_{2} O = 4.18 JK^{- 1} g^{- 1}

Density of

H_{2} O = 1.0 gcm^{- 3}

Assume no change in volume of solution on mixing]

Answer: 82

The acid supplies

H^{+} = 0.400 \times 0.2 \times 2 = 0.16

mol and the base supplies

O H^{-} = 0.600 \times 0.1 = 0.06

mol, so hydroxide is in deficit and only

0.06

mol of water is formed. The heat released is

q = 0.06 \times 57.1 = 3.426 kJ = 3426 J

The mixture occupies

400 + 600 = 1000

mL and at unit density has a mass of

1000

g, so

Δ T = \frac{3426}{1000 \times 4.18} = 0.82 K

The temperature of the final solution rises by

82 \times 1 0^{- 2}

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Integer answer