Rob Gronkowski Earns AFC Offensive Player Of The Week After Torching Steelers

Rob Gronkowski was recognized Wednesday as the AFC’s Offensive Player of the Week after enjoying one of the best games of his career in the New England Patriots’ Week 15 win over the Pittsburgh Steelers.

After being suspended for the previous game, Gronkowski caught nine passes on 13 targets Sunday for a career-high 168 yards, including three receptions for 69 yards on the drive that produced New England’s game-winning touchdown. The tight end also found the end zone on a pivotal two-point conversion to help the Patriots secure a 27-24 victory at Heinz Field.

“It was easily the best tight end performance I’ve ever seen live,” fellow Patriots tight end Dwayne Allen said after the game.

In his last two games, Gronkowski has 18 catches on 23 targets for 315 yards. He was one of four Patriots players to earn a Pro Bowl selection Tuesday, along with quarterback Tom Brady, fullback James Develin and special teamer Matthew Slater.

New England will host the Buffalo Bills on Sunday at Gillette Stadium.

Thumbnail photo via Philip G. Pavely/USA TODAY Sports Images

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